Remove denominators in de Rham cohomology
I finally found a counter example in a note of Émile Picard written in 1899.
It is an algebraic example but it is easily translated into a rational example with an extra variable. Let $P(t)$ be a square-free polynomial of degree at least three. There is a non-zero polynomial $U(x,y)$ such that $$ \partial_x\left( \frac{\sqrt{P(x)}}{(y-x)\sqrt{P(y)}}\right) - \partial_y\left( \frac{\sqrt{P(y)}}{(x-y)\sqrt{P(x)}}\right) = \frac{U(x,y)}{\sqrt{P(x)}\sqrt{P(y)}}.$$ Let $F$ be the right-hand side. Note the pole $(y-x)$ which appears inside the derivatives but not in $F$.
Picard proved that there is no rational functions $G_x$ and $G_y$ in $x$, $y$ and $\sqrt{P(x)}\sqrt{P(y)}$ such that:
$F = \partial_x G_x + \partial_y G_y$
$G_x$ and $G_y$ have no pole outside those of $F$
Indeed, this would imply that $\oint F \mathrm{d}x \mathrm{d}y$ is zero on every cycle on the Riemann variety of $\sqrt{P(x)}\sqrt{P(y)}$. However, let $a_1$, $a_2$ and $a_3$ be three distinct roots of $P$, and let $\gamma_1$ be a contour in $\mathbb C$ which encloses $a_1$ and $a_2$ but no other root of $P$, and let $\gamma_2$ be a contour in $\mathbb C$ which encloses $a_2$ and $a_3$ but no other root. $\gamma_1 \times \gamma_2$ induces a cycle on the Riemann variety of $\sqrt{P(x)}\sqrt{P(y)}$, and we can check that $$\oint_{\gamma_1\times \gamma_2} F(x,y)\mathrm{d}x \mathrm{d}y = 4i\pi.$$
Thanks M. Picard !
Picard, Émile. (1899). Quelques remarques dur les intégrales doubles de seconde espèce dans la théorie des surfaces algébriques, Comptes rendus hebdomadaires des séances de l'Académie des sciences, 129, 539–540, Gallica
Since this question is back on the front page, here is another counterexample. Again, it is algebraic but I think it could be made rational in one more variable; it has the nice feature that all the Hodge structures involved are of Tate type.
Let $f(z) = \prod_{i=1}^d (z-z_i)$ with $z_1$, $z_2$, ..., $z_d$ distinct complex numbers. Let $X$ be the affine surface $$xy = f(z)$$ and let $Z = \{ y=0 \}$. Then $X \setminus Z$ projects isomorphically to $\{ (y,z) : y \neq 0 \}$, and we see that $H^1(X \setminus Z) \cong \mathbb{C}$ and $H^2(X \setminus Z) = 0$. However, $Z$ is the disjoint union of $d$ copies of $\mathbb{C}$. So $H^1(X \setminus Z) \to H^0(Z)$ is not surjective, and $H^2(X) \cong \mathbb{C}^{d-1}$ does not inject into $H^2(X \setminus Z)$.
Concretely, let $\gamma$ be a path from $z_i$ to $z_j$ in $\mathbb{C}$, not passing through another root of $f$. Let $$S = \{ (x,y,z) : |x|=|y|=\sqrt{|f(z)|},\ xy=f(z),\ z \in \gamma \}.$$ So $S$ is a $2$-sphere.
On $X$, we have $dx/x + dy/y = f'(z)/f(z) dz$ and thus $(dx \wedge dz)/x = -(dy \wedge dz)/y$, wherever these expressions are defined. Now, $(dx \wedge dz)/x$ is defined except where $x=0$ and $-(dy \wedge dz)/y$ is defined except where $y=0$, and the locus where $x=y=0$ is codimension $2$ in the smooth variety $X$, so $(dx \wedge dz)/x$ extends to a global $2$-form on $X$. For any polynomial $g(z)$, we have $\int_S g(z) (dx/x) \wedge dz = (2 \pi i) \int_{z_i}^{z_j} g(z) dz$.
So, for most choices of $g$, we have $\int_S g(z) (dx/x) \wedge dz \neq 0$, and $g(z) (dx/x) \wedge dz$ is a closed, non-exact, $2$-form.