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.
I think that the answer to your question is no.
In order to construct a counter-example, the idea is the following: let $T$ be a Kahler current, and $E$ a $d$-closed positive current on a compact complex $3$-fold $X$ such that $T^3>0$ and $E^3<0$. Then, for $t>0$, the current $S_t:=T+tE$ is a Kahler current. Under the above conditions, you can find $t>0$ such that $S_t^3=0$. Indeed, $S_t^3=T^3+3tT^2E+3t^2TE^2+t^3E^3$ and for $t=0$, $S_0^3=T^3>0$ while for $t\to \infty$, $S_t^3<0$, so there is a $t>0$ so that $S_t^3=0$.
Now, for this $t$, denote by $\tau_t$ a smooth representative of the class of $S_t$. Then $\tau_t^3=0$. Now for instance, for $q=n=3$, the map $\tau_t^3\wedge:H^0(X,\Omega^0)\to H^3(X,\Omega^3)$ is the zero map, while $H^0(X,\Omega^0)$ and $H^3(X, \Omega ^3)$ are $1$-dimensional.
In order to fulfill these conditions, you can take $Y$ to be Hironaka's example, it is a modification of $CP^3$, so there is a map $p:Y\to CP^3$, and denote by $\omega$ the FS metric on $CP^3$
Then the $X$ mentioned above is the blow up of $Y$ at a point $x$, denote by $\pi:X\to Y$ the blow-up. If $E$ is the exceptional divisor, then $[E]$ is a positive current, and $E^3=-1<0$, and $T$ is $\pi^*p^*\omega+\varepsilon S$, where $S$ is some Kahler current on $X$ ans $\varepsilon$ is such that $T^3>0$. Then, as mentiond above, you can find $t>0$ such that $T+t[E]$ has zero self-intersection.
Best Answer
At least when $M$ is compact, which I assume here, the standard reference for this subject is probably the book
but the following books are excellent references as well :
A proof could run as follows:
For an explicit construction of $L$: the standard proof of Hodge's theorem is cohomological, and uses a long exact sequence in cohomology. Tracing down the arguments gives an actual construction if one is given an open covering of $M$ by contractible open sets.
The arguments really require that $\omega$ be integral. A generic complex torus of dimension $>1$ carries no complex line bundle, but has many Kähler forms.