The answer is 'no, not necessarily'.
Consider the following example: Let $M=N=\mathbb{CP}^2$, let $(\omega_0,J_0)$ be the standard Fubini-Study Kähler structure on $M$. Now let $f$ be an arbitrary, but '$C^2$-small' smooth function on $M$, so that $\omega_0 + t\,\mathrm{i}\,\partial\bar\partial f$ is nondegenerate (and hence symplectic) for all $0\le t\le 1$.
Let $\omega_M = \omega_0 + \mathrm{i}\,\partial\bar\partial f$ and let $J_M= J_0$. Let $\omega_N=-\omega_M$ and let $J_N=-J_0$. Note that $g_M=g_N$, so $(M,g_M)$ and $(N,g_N)$ are isometric via the identity map.
When $f$ is chosen sufficiently generically, the isometry group of $g_M$ will consist of only the identity, so suppose this.
Note that $(M,J_M)$ and $(N, J_N)$ are biholomorphic, since $\mathbb{CP}^2$ is biholomorphic to its conjugate complex manifold.
Note that, by Weinstein's theorem, since $\omega_M$ and $\omega_0$ are $C^0$-close and obviously cohomologous, there is a symplectomorphism between $(M,J_M)$ and $(M,\omega_0)$. Similarly, there is a symplectomorphism between $(N,J_N)$ and $(N,-\omega_0)$. Since, as has already been noted, $(M,\omega_0)$ and $(N,-\omega_0)$ are symmplectomorphic, it follows that $(M,\omega_M)$ and $(N,\omega_N)$ are symplectomorphic.
However, when $f$ is chosen sufficiently generically, the only map $\psi:M\to N$ that aligns the metrics $g_M$ and $g_N$ is the identity, which is neither a biholomorphism nor a symplectomorphism.