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
If $X'$ is a Mukai flop of a compact hyper-Kähler manifold $X$, then $X'$ is in Fujiki class $\mathcal{C}$ and carries a holomorphic symplectic form $\sigma$. Taking the real part or the imaginary part of $\sigma$ gives a symplectic form on $X'$.
There exist however Mukai flops which are not Kähler, see e.g. this paper of Yoshioka, Section 4.4.