I understand the terms as I wrote in my comment. Also, judging by the tags, I assume that everything (the manifold and the action) is smooth. Then the answer is as follows.

In general, no. Let $G$ be the group of all upper-triangular matrices with positive diagonal entries. It acts on $\mathbb R^2$ as a subgroup of $GL(2,\mathbb R)$. Consider $x=(1,0)$. Its orbit is the coordinate ray $\{(t,0):t>0\}$. Its stabilizer $G_x$ consists of matrices whose upper-left element is 1 and the second column is arbitrary. This stabilizer acts transitively on the upper half-plane, so there are no invariant transversals to the horizontal line.

If $G$ is compact, then yes. By compactness, there is a Riemannian metric on $X$ invariant under $G$. Let $Z$ be the orthogonal complement to $T_x\mathcal O$ in $T_xX$ (with respect to the Riemannian scalar product). Let $B$ be a small open ball in $Z$ (centered at the origin). Then the submanifold $\exp_x(B)$, where $\exp_x$ is the Riemannian exponential map, is invariant under $G_x$.

In general, no. Let $G$ be the group of all upper-triangular matrices with positive diagonal entries. It acts on $\mathbb R^2$ as a subgroup of $GL(2,\mathbb R)$. Consider $x=(1,0)$. Its orbit is the coordinate ray $\{(t,0):t>0\}$. Its stabilizer $G_x$ consists of matrices whose upper-left element is 1 and the second column is arbitrary. This stabilizer acts transitively on the upper half-plane, so there are no invariant transversals to the horizontal line.

If $G$ is compact and everything is smooth, then yes. By compactness, there is a Riemannian metric on $X$ invariant under $G$. Let $Z$ be the orthogonal complement to $T_x\mathcal O$ in $T_xX$ (with respect to the Riemannian scalar product). Let $B$ be a small open ball in $Z$ (centered at the origin). Then the submanifold $\exp_x(B)$, where $\exp_x$ is the Riemannian exponential map, is invariant under $G_x$.