The answer is no for $\mathbb F_2$. A minimal invertible counterexample is $M = \left[\begin{matrix}1 & 1 & 0\\0 & 1 & 0\\0 & 1 & 1\end{matrix}\right]
$. This can be verified by observing that all orthogonal $3 \times 3$ matrices over $\mathbb F_2$ are permutation matrices. So if you multiply $M$ by each of the six permutation matrices, you never get a symmetric matrix.

The fact that $O_3(\mathbb F_2)$ only contains permutation matrices can be verified using the "inner-product preserving" property of orthogonal matrices.

Computational experiments show that a counterexample exists for $\mathbb F_3$ of size $3 \times 3$. For $\mathbb F_5$, there is a $2 \times 2$ counterexample.

Regarding alternatives to polar decomposition, in Gutin - Generalisations of singular value decomposition to dual-numbered matrices [1] I introduce an analogue of SVD for the ring of dual numbers equipped with the involution $a + b \varepsilon \mapsto a - b\varepsilon$. This analogue of SVD exists for every dual-numbered matrix, while the polar decomposition (its obvious generalisation) does not. What this means is that when generalising SVD it might be profitable to allow the matrix $S$ in $USV^*$ to not necessarily be Hermitian.

[1] - https://www.tandfonline.com/doi/full/10.1080/03081087.2021.1903830, but this has some formatting errors, so instead see https://arxiv.org/abs/2007.09693