Let M be an open 2n dimensional manifold that admits a non-degenerate 2 form. By Gromov's h-principle we know that M admits a symplectic form and that the homotopy type of the space symplectic forms on M is homotopic to the space of non-degenerate two forms on M, which in turn is homotopic to the space of all almost complex structures $\mathscr{J}$ on M.
For what open manifolds do we understand the topology of $\mathscr{J}$? As explained here for contractible manifolds M, $\pi_n(\mathscr{J}) = \pi_n ( GL(2n,\mathbb{R})/GL(n,\mathbb{C}))$.
In particular, $\mathscr{J}$ has two connected components when M is contractible. What do we know about $\pi_0(\mathscr{J})$ for other non-contractible open manifolds?
One observation is that $\pi_0(\mathcal{J})$ can be infinite.
Two almost complex structures in the same path component of $\mathcal{J}$ have the same Chern classes, so for each $i$, we have well-defined maps $c_i : \pi_0(\mathcal{J}) \to H^{2i}(M; \mathbb{Z})$. By definition, the image is precisely the set of cohomology classes which can be realised as $c_i(TM, J)$ for some choice of almost complex structure $J$ on $M$.
Example: An orientable six-manifold $M$ admits an almost complex structure $J$ with $c_1(TM, J) = \alpha$ if and only if $\alpha \equiv w_2(M) \bmod 2$. In particular, if $M$ is spin, then $\alpha$ is realised if and only if it is divisible by $2$. For $M = \mathbb{CP}^3\setminus\{p\}$ for example, after choosing an isomorphism $H^2(\mathbb{CP}^3\setminus\{p\}; \mathbb{Z}) \cong \mathbb{Z}$, we see that the image of $c_1 : \pi_0(\mathcal{J}) \to H^2(\mathbb{CP}^3\setminus\{p\}; \mathbb{Z}) \cong \mathbb{Z}$ is $2\mathbb{Z} \subset \mathbb{Z}$. In particular, $\pi_0(\mathcal{J})$ is infinite.
In general, the maps $c_i : \pi_0(\mathcal{J}) \to H^{2i}(M; \mathbb{Z})$ are not injective (even if we restrict to the space of almost complex structures inducing a given orientation).
If you're willing to consider compact manifolds, then the paper Topology of Almost Complex Structures on Six-Manifolds by Granja and Milivojevic may be of interest to you.