Let's say that there is a symplectic manifold $(M,\omega)$ with condition of $[\omega / 2\pi ]\in H^2(M;\mathbb{Z})$.
Then in what condition can I get a complex line bundle $L\twoheadrightarrow M$ in what condition?
Does it have to satisfy condition of $c_1(L)=[\omega/2\pi]$? Here, $c_1$ is chern class.
John is right, if you are not asking for any particular properties, there are no conditions: trivial bundles are always possible.
Now, if you are interested in Chern classes and symplectic manifolds, you probably want some explanation about prequantum line bundles.
Considering tensor products of vector bundles and bundle isomorphisms, the group of equivalence classes of complex line bundles over a manifold $M$ is isomorphic to $\check{H}^2(M;\mathbb{Z})$ (nice exercise, you should try this).
A inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ induces a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow \check{H}^2(M;\mathbb{R})$ and under a identification $\check{H}^2(M;\mathbb{R})\cong H_{dR}^2(M;\mathbb{R})$ one has a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow H_{dR}^2(M;\mathbb{R})$. Therefore, given a fixed closed form $\omega\in\Omega^2(M;\mathbb{R})$ whose de Rham class $[\omega]$ lies in the image of the homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow H_{dR}^2(M;\mathbb{R})$, there are equivalence classes of complex line bundles with hermitian connexions such that their curvature is $-i\omega$. Note that the map $\check{H}^2(M;\mathbb{Z})\longrightarrow H_{dR}^2(M;\mathbb{R})$ has a kernel.
You should not write, though, that the class of your de Rham form belongs to $\check{H}^2(M;\mathbb{Z})$, that is not the case (and it does not make any sense): to be integral really means to lie in the image of the aforementioned homomorphism.
P.D.: $2\pi$'s, $i$'s, and signs conventions are all different in the literature, just pick one that you like.