Finding a 1-form adapted to a smooth flow
If I understand correctly, there is already a counterexample on the torus:
On the $xy$-plane $\mathbb{R}^2$, let $X$ be the vector field $$ X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\partial\ }{\partial y}. $$ Now let $T^2=\mathbb{R}^2/\Lambda$ where $\Lambda$ is the lattice generated by $(2\pi,0)$ and $(0,2\pi)$. Since $X$ is invariant under this lattice, it gives rise to a well-defined, nowhere vanishing vector field on $T^2$, which I will also call $X$. It has two closed orbits $C_0$ defined by $x\equiv0\mod 2\pi$ and $C_1$ defined by $x\equiv \pi\mod 2\pi$, while every other orbit is nonclosed and has $C_0$ as its $\alpha$-limit and $C_1$ as its $\omega$-limit. In particular, the only functions on $T^2$ that are constant on the $X$-orbits are the constant functions.
Now suppose that $\theta$ is a $1$-form on $T^2$ such that $\mathcal{L}_X\theta$ is exact. Then, by Cartan's formula, $$ \mathrm{d}\bigl(\theta(X)\bigr) + i(X)(\mathrm{d}\theta) = \mathrm{d}h $$ for some function $h$ on $T^2$. I.e., $$ i(X)(\mathrm{d}\theta) = \mathrm{d}\bigl(h-\theta(X)\bigr), $$ implying that the function $h-\theta(X)$ is constant on the flow lines of $X$ and hence must be constant. Thus, $i(X)(\mathrm{d}\theta)=0$. Since $X$ is nowhere vanishing, $\mathrm{d}\theta$ must vanish identically, so $\theta$ must be closed.
Now, the integrals of $\theta$ over the two closed orbits (oriented so that $\mathrm{d}y>0$, say) must be equal, since, oriented this way, they are homologous in $T^2$. However, $X$ orients $C_0$ and $C_1$ so that they are opposite in homology, so the integrals using $X$ to orient them positively must have opposite signs (or vanish). Hence, it cannot be that $\theta(X)>0$ everywhere.
Added on July 29: A request has been made for an example of such a vector field without any closed leaves. This is easy to provide, as follows:
Let $\mathbb{T}^3 = \mathbb{R}^3/(2\pi\mathbb{Z}^3)$ be the 'square' $3$-dimensional torus, i.e., $xyz$-space where $x$, $y$, and $z$ are $2\pi$-periodic. Let $$ X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\partial\ }{\partial y} + \sqrt 2\,\cos x\,\frac{\partial\ }{\partial z}\,, $$ which is a well-defined vector field on $\mathbb{T}^3$. The $2$-tori $C_0$ (defined by $x\equiv0\,\mathrm{mod}\,2\pi$) and $C_\pi$ (defined by $x\equiv\pi\,\mathrm{mod}\,2\pi$) are invariant under $X$ and all the flow-lines of $X$ in $C_0$ and $C_\pi$ are dense in their respective $2$-tori. Meanwhile, every other flow-line of $X$ $\alpha$-limits to $C_0$ and $\omega$-limits to $C_1$. In particular, any function on $\mathbb{T}^3$ that is constant on the flow-lines of $X$ is necessarily constant. Just as above, it follows that if $\theta$ is a $1$-form on $\mathbb{T}^3$ that is adapted to $X$, then $i(X)(\mathrm{d}\theta)=0$. It follows that there is a smooth function $\lambda$ on $T$ such that $$ \mathrm{d}\theta = \lambda\,i(X)(\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z) = \lambda\,\bigl(\sin x\,\mathrm{d}y\wedge\mathrm{d}z + \cos x\, \mathrm{d}z\wedge\mathrm{d}x + \sqrt2\,\cos x\,\mathrm{d}x\wedge\mathrm{d}y \bigr) $$ Taking the exterior derivative of both sides of this equation yields the identity $$ 0 = \mathrm{d}\lambda(X) + \lambda\,\cos x\,. $$ Consequently, $$ \mathrm{d}(\lambda\,\sin x)(X) = 0 $$ Thus, $\lambda\,\sin x$ is constant along the flow lines of $X$ and hence is constant. Since it vanishes on $C_0$ and $C_\pi$, the function $\lambda\,\sin x$ must vanish identically. Hence $\lambda$ vanishes identically, i.e., $\mathrm{d}\theta = 0$.
Since $\theta$ is closed on $\mathbb{T}^3$, its integral over any two homologous closed oriented curves must be equal. However, just as in the $2$-dimensional case, using the hypothesis that $\theta(X)>0$, one can easily construct a closed oriented curve $\gamma_0$ in $C_0$ on which $\theta$ pulls back to be positive while its translate by $\pi$ in the $x$-direction, say $\gamma_\pi$ lies in $C_\pi$ and has the property that $\theta$ pulls back to $\gamma_\pi$ to be negative, making it impossible for the integrals of $\theta$ over the two curves to be equal.
If the vector field is geodesible then such $1$ - form exists.
A geodesible flow on a manifold $M$ is a one dimensional foliation associated with a non vanishing vector field such that all leaves of the foliation are geodesics for some Riemannian metric on $M$. This is mentioned in the book by Philippe Tondeur, Geometry of Foliations, Page 71 Proposition 6.8.
I quote the Proposition here:
6.8 Proposition: Let $V$ be a non singular vector field on $M$. Then the following conditions are equivalent:
(i) The flow of $V$ is geodesible,
(ii) there exist a $1$-form $\chi\in \Omega^1(M)$ such that $\chi (V)>0$ and $i(V)d\chi=0$
Another equivalent condition: From the Proposition 6.7 of the same page of the above book we learn that a non vanishing vector field $V$ is geodesible if and only if there is a subbundle $E$ of $TM$, complementary to $V$, such that for every vector field $X$ tangent to $E$ we have $[X,V]$ is tangent to $E$. This obviously implies the following:
Obvious Fact: Assume that $U$ is an open subset of $\mathbb{C}$ and $V: U \to \mathbb{C}$ is a non vanishing holomorphic function. Then the vector field on $U$ determined by $V$ generates a geodesible foliation. The reason is that in the above equivalent formulation on can put $E=$ the direction determined by $W=iV$. Note that $[V, fW]=(V.f)W$ since $[V,W]=0$.
In the following "Edited" part, we explain about a possible relation between the concept "geodesible flow" and the problem of the number of limit cycles of a polynomial vector field:
EDITED:
This very interesting answer contains an example of a non geodesible foliation of $\mathbb{R}^2 \setminus \{0\}$ associated with a polynomial vector field of degree $5$.
But in degree $2$, the situation is different. The key point is that every limit cycle of a quadratic vector field is necessarily a convex curve. We explain in details as follows::
Put $$\begin{cases} x'=P(x,y)\\y'=Q(x,y) \end{cases}\;\;\;\; (V)$$
where $P.Q$ are polynomials of degree $2$ with $P(0,0)=Q(0,0)=0$. It is well known that every closed orbit of $(V)$ is a convex curve.( See 4.6,page 108 of this book or see directly theorem $1$ of this paper).
This shows that if a limit cycle $\gamma$ of $(V)$ surrounds origin then it does not intersect the algebraic curve $$C=\{(x,y)\mid yP-xQ=0\}$$ This is an immediate consequence of the following fact:
The limit cycle $\gamma$ is a convex closed curve and surrounds origin. Thus no point $(x,y)$ of $\gamma $ satisfy $yP-xQ=0$.
Now put $$\chi=\frac{1}{x^2+y^2}(ydx-xdy)$$ then $\chi$ is a closed $1$_form on the punctured plane which satisfies $\chi(V) \neq 0$ on $\mathbb{R}^2 \setminus C$.
In fact $\chi= d \theta$ where $\theta$ is the standard "argument" in the polar coordinate $(r, \theta)$ ( My serious thanks to Ben MacKay for his valuable suggestion of $d\theta$ as a possible candidate to find a $1$-form $\chi$ which (globally ) satisfies the above Theorem of Sullivan, that is global satisfaction of $\chi (V) \neq 0$ ). But in reality $\chi=d\theta$ does not satisfies $\chi (V) \neq 0$ golabally on the phase space , since it vanishes on $C$. But, fortunately, no limit cycle of a quadratic vector field as $(V)$ can intersect this obstruction curve $C$. However $C$ need not be necessarily a transversal curve!(It may be tangent to the vector field $(V)$ at some points).
So with $(V)$ and $C$ described above, we summarize the above statements as follows:
Every quadratic vector field $(V)$ is geodesible on $\mathbb{R}^2 \setminus C$. A limit cycle which surrounds origin, it never intersect $C$.
So if we can control the sign of the curvature of the corresponding Riemannian metrics which make the flow of $(V)$ geodesible, then we can count the number of limit cycles of $(V)$ which surround origin. But if a limit cycle surround a singularity other than origin, we translate the singularity to the origin. This procedure would count all possible limit cycles of an arbitrary quadratic vector field.
In the above, by "Control of sign of the curvature" we mean that we need that the following situation would occur :
Either the curvature is identically zero, corresponding to the center singularity, or the curve consisting of all points with zero curvature would be a transversal curve, hence no limit cycle can intersect it. This is explained here:
Limit cycles as closed geodesics(in negatively or positively curved space)
As an updated related post please see the following
Limit cycles as closed geodesics(2)
For a slightly different perspective: The Reeb foliation of an annulus (https://en.wikipedia.org/wiki/Reeb_foliation) can be doubled to give a flow on a torus, generated by a nowhere zero vector field $X$, such that all but one trajectory spirals towards a closed loop in forward time. This flow has the property that (i) all but 2 trajectories are nonrecurrent and (ii) the foliation has no closed transversal (it is the standard example of a "non-taut" foliation, see e.g. Thurston, "A Norm on the Homology of a 3-Manifold").
It follows that there is no $X$-invariant 1-form $w$ with $w(X)>0$. Indeed, if $dw \neq 0$ then $|dw|$ gives a smooth X-invariant measure, contradicting (i), and if $dw = 0$ then $w$ is a closed, nowhere zero 1-form, and the leaves of the measured foliation defined by $Ker(w)$ are recurrent and transverse to the leaves of $X$, giving a closed transversal, contradicting (ii).
This is the same as Bryant's example; as he observes, one can uses the additional fact that every $X$-invariant function is constant to even rule out $w$ with $L_X(w) = df$.