Almost complex manifolds are orientable

Here's a slightly different way of looking at it. First, note that we can choose a Riemannian metric $g$ on $M$ that satisfies $g(X,Y) = g(JX, JY)$ (start with any Riemannian metric $h$ and then define $g(X,Y) = h(X,Y) + h(JX, JY)$). You can then show that $\omega(X,Y) = g(X,JY)$ is skew-symmetric:

\begin{align} \omega(X,Y) & = g(X,JY) = g(JY, X) \\& = g(J^2Y, JX) = -g(Y, JX) = -\omega(Y,X) \end{align}

and is therefore a 2-form. $\omega$ is also non-degenerate, since $\omega(X, -) = - g(JX, -)$ and a metric is non-degenerate by definition. Then since $\omega$ is non-degenerate, its top exterior power $\omega^n = \omega\wedge \dots \wedge \omega$ is nowhere zero. A nowhere vanishing top form defines an orientation on $M$, so we are done.

I see you have added an answer - and Charlie's is fine too. Here is another way...

First of all - you have checked this, but for completeness - if we endow a real vector space $V$ of dimension $2n$ with a complex structure (i.e.,view it as a $C=\mathbb R[J]$ module), then we have chosen an orientation. For, suppose $v_1,\cdots, v_n$ and $w_1,\cdots, w_n$ are $C$-bases, and $g $ is the transition matrix ($g$ has entries in $C$, or $g = a + Jb$, with $a$, $b$ real) between the two, with determinant $d\in C$. Then $g,$ viewed as a transformation over the reals, has determinant $\det_{\mathbb R}g =d\bar d$, and hence is positive.

Proof: We wish to calculate $\det_{\mathbb R} g$. Tensor $V$ with $\mathbb C$ - that is, extend scalars. The determinant does not change:$${\det}_{\mathbb R}g = {\det}_{\mathbb C} (g\otimes 1).$$

So we wish to calculate $\det_{\mathbb C} (g\otimes 1)$: $g\otimes 1$ preserves $V_{\pm i}$, the $\pm i$-eigen-spaces of $J$. Thus, since $g\otimes 1$ acts as $a\pm ib$ on $V_{\pm i}$, and $V\otimes {\mathbb C}=V_i\oplus V_{-i}$, $${\det}_ {\mathbb R} g=\det g\otimes 1 = \det g\otimes 1|_{V_i}\cdot \det g\otimes 1|_{V_{-i}} = \det (a +ib) \cdot \det (a-ib) >0.$$

Therefore, since $$ w_1\wedge\cdots \wedge w_n \wedge J w_1\wedge\cdots \wedge J w_n = {\det}_ {\mathbb R} g\ v_1\wedge\cdots \wedge v_n \wedge J v_1\wedge\cdots \wedge J v_n,$$ the orientation is well-defined.

To answer the question proper...

By assumption, $J$ is a global tensor, and thus each $T_p(M)$ is endowed with a $J$ structure, and hence (by the previous argument), an orientation, independent of charts. To bring this in line with the original formulation in your question: suppose $$\phi: U\subset M\to \mathbb R^{2n},$$ is a chart, and $p\in U$. Then $\partial/\partial x_1|_p,\cdots ,\partial/ \partial x_{2n}|_p$ - obtained from $\phi$ (or $\phi^{-1}$), is an oriented basis. If the orientation fails to match the $J$ orientation, modify $\phi$ by swapping the first two coordinates. By continuity of $J$, the (modified, if needed) chart will give you an oriented frame on all of $U$ that matches that of $J$. The manifold $M$ is therefore orientable - that is the transition function jacobian determinants will be positive.

I finally figured this out, so I want to post my own answer, since I couldn't find a satisfactory answer myself online.

It suffices to prove that for every $p\in M$, there exists a local frame $(\sigma_1,\cdots,\sigma_n,J\sigma_1,\cdots,J\sigma_n)$ in some open neighborhood $U$ of $p$.

Fix $p\in M$ and choose an ordered basis $(v_1,\cdots,v_n,Jv_1,\cdots,Jv_n)$. Choose sections $\sigma_1,\cdots,\sigma_n$ such that $\sigma_i(p)=v_i$ for each $1\leq i\leq n$. Since $(\sigma_1|_p,\cdots,\sigma_n|_p,J\sigma_1|_p,\cdots,J\sigma_n|_p)$ is linearly independent,

$$\omega=\sigma_1\wedge\cdots\sigma_n\wedge(J\sigma_1)\wedge\cdots\wedge(J\sigma_n)\neq0$$

at $p$. Since $\omega$ is continuous, it is nonzero for some neighborhood $U$ of $p$. In other words,$(\sigma_1,\cdots,\sigma_n,J\sigma_1,\cdots,J\sigma_n)$ is linearly independent in $U$, so that it is a local frame.