Suppose that we have a complex manifold $X$, and a line bundle $L$ over $X$. It is known that the line bundles over $X$ are parametrized by their Chern class, the Chern class being the cohomology class of the curvature of a connection on $L$, which must be integral. Thus $L$ admits a connection with curvature $\omega$ iff $[\omega]$ is an integral cohomology class.  

My question is this: what if we are interested instead in complex connections on $L$ (connections that are $\mathbb{C}$-linear instead of $\mathbb{R}$-linear)? Given a complex (2,0)-form $\omega$ on $X$, under what circumstances will $L$ admit a complex connection with curvature $\omega$?



Line bundles with complex connection
If I understood this right, the answer to your original question is somewhat trivial. You may check that if a complex connection $\nabla$ on $L$ has curvature $\omega$, then the cohomologous $2$-form $\omega+d\alpha$ is the curvature of the connection $\nabla+\alpha$. Hence any closed $2$-form representing integral class can be the curvature form of a complex connection on a line bundle (which just means a $C^\infty$ vector bundle, not holomorphic one).


However, there is a version of your question which admits a non-trivial answer. 


If you take $L$ to be a **holomorphic** line bundle and endow it with a Hermitian metric then there is an unique Hermitian $(1,0)$-connection called the Chern connection, its curvature form $\omega$ is a real-valued (1,1)-form (real-valued means that $\omega=\bar{\omega}$) representing an integral class. 


A non-trivial result is that the converse is also true when $X$ is compact Kähler, i.e., any real-valued (1,1)-form representing an integral class is the curvature form of the Chern connection on a Hermitian holomorphic line bundle. This is a consequence of the $\partial\bar\partial$-lemma and the Lefschetz theorem on $(1,1)$-classes. 


