I have a partial answer to my own question that I posed 5 days ago. It turns out that at least in the case of a smooth symplectic manifold you can perform an orbit type stratification. Here is why:

Asume $X$ is a symplectic manifold equipped with a symplectic form $\omega$. Assume furthermore that a compact Lie group acts by symplectomorphisms on $X$. Then if we ignore the symplectic form for a while, $X$ is a smooth $G$-manifold. It therefore naturally decomposes in so-called orbit types $X_{(H)}:=\{x\in X: (Stab(x))=(H)\}$, where $H$ is a closed Lie subgroup of $G$ and the brackets $(~~)$denote the conjugacy class. These are in general not connected, but their connected components, $X_{(H)}^{i}$, have the structure of locally closed submanifolds. These connected components are referred to as *orbit type strata* of $X$ and satisfy a bunch of useful conditions, the most important of which is the *frontier condition*. These conditions are irrelevant here, so I will not mention them further.
The smooth manifold $X$ is decomposed in a disjoint union of orbit type strata.
Interestingly, each orbit type stratum $X_{(H)}^{i}$, coincides with some connected component $X_{H}^{j}$ of the isotropy type $X_{H}:=\{x\in X| Stab(x)=H\}$. Here is where the symplectic picture comes into play. By Lemma 27.1 in the book "*Symplectic techniques in physics*" by Guillemin and Sternberg the tangent space at the every point $x$ in $X_{H}$, $T_xX_H$ is a symplectic subspace of the symplectic tangent space at the point $x$ in all of $X$, $(T_xX, \omega_x)$.
Hence $X_H$ is in fact a symplectic sub-manifold of $X$. Hence each connected component $X_{H}^{j}$ will be a symplectic submanifold. Thus, the orbit types $X_{(H)}^{i}$ will be symplectic submanifolds. The additional conditions, that strata need to satisfy, are topological in nature so I don't believe that the symplectic form $\omega$ will affect them. In light of what I tried to explain in the above, I believe that a smooth symplectic manifold $X$ endowed with a group action by a compact symplectic Lie group, can be stratified by symplectic orbit types.
Moreover, I am quite certain that the above argumentation goes through in the case of a complex symplectic manifold, since in that case we have in advance a complex stratification by holomorphic orbit types. The only thing missing is that the strata admit a closed, non-degenerate $(2, 0)$-form. This however can be demonstrated in a similar fashion as in the book of Guillemin and Sternberg.

10more comments