This is not necessarily true, even locally. Consider $M \subset \mathbb{R}^6$ given by the graph of the function $f : \mathbb{R}_x^3 \rightarrow \mathbb{R}_y^3$ with

$$f(x_1,x_2,x_3) = (x_2x_3, 0, -x_1x_2)$$

Then at a point $p = (\vec{x},f(\vec{x})) \in M \subset \mathbb{R}_x^3 \times \mathbb{R}_y^3$, we compute $T_pM$ is given by the span of the vectors

$$e_1 = \partial_{x_1} -x_2\partial_{y_3}$$
$$e_2 = \partial_{x_2} + x_3\partial_{y_1} - x_1\partial_{y_3}$$
$$e_3 = \partial_{x_3} + x_2\partial_{y_1}$$

Explicitly, we have that in this basis,

$$\omega = \begin{pmatrix} 0 & -x_3 & -x_2 \\ x_3 & 0 & -x_1 \\ x_2 & x_1 & 0 \end{pmatrix}$$

from which it is easy to compute that so long as $p \neq 0$, any $X$ such that $\omega(X,v) = 0$ for all $v \in T_pM$ (i.e. $X$ is in the nullspace of the above matrix) must satisfy that $$X(p) \in \mathrm{span}\left\langle\begin{pmatrix}x_1 \\ - x_2 \\ x_3\end{pmatrix}\right\rangle$$ By continuity, $X(0)$ would need to be a multiple of $e_1$ (by looking at $X(p)$ along the axis $x_2 = x_3 = 0$) and also a multiple of $e_3$ (similarly). So there can be no nonzero $X(p)$ in a neighborhood of $0$.

For $M$ in higher dimensions, one can repeat the argument by successively taking $M \times \mathbb{R}^2 \subset \mathbb{R}^{2n} \times \mathbb{R}^4$, where the $\mathbb{R}^2 \subset \mathbb{R}^4$ we choose is symplectic. For a compact $M$, simply choose one which matches this model near $0$.