Let's assume that $X$ admits a $K$-point $x$ and use the corresponding geometric point as the base point. The existence of a rational point is in fact necessary for a positive answer, as explained by S. carmeli.

In terms of etale fundamental groups the question can be paraphrased as follows: given an open subgroup $H\subset \pi_1(X_{\overline{K}},x)$ does there exist an open subgroup $H'\subset H$ such that the action of the Galois group $G_K$ on $\pi_1(X_{\overline{K}},x)$ preserves $H'$.

This is true and follows from $\pi_1(X_{\overline{K}},x)$ being topologically finitely generated. Consider the subgroup $\Gamma_H\subset G_K$ consisting of elements $\gamma\in G_K$ such that $\gamma(H)=H$. Let $h_1,\dots, h_n$ be a set of topological generators of $H$ ($H$ is topologically finitely generated because it has finite index in $\pi_1(X_{\overline{K}})$). Then $\Gamma_H$ can be expressed as $\{\gamma\in G_K|\gamma(h_i)\in H\}$ so $\Gamma_H$ is an intersection of finitely many open subset, hence is an open subgroup. In particular, $\Gamma_H$ has finite index in $G_K$. Take $\Gamma\subset \Gamma_H$ to be an open subgroup which is moreover normal in $G_K$.

Let $g_1,\dots, g_m$ be a set of representatives of cosets of $\Gamma$ in $G_K$. Then $H'=\bigcap g_i(H)$ is an open subgroup with the desired property. Indeed, suppose that $x\in H'$ and $\gamma g_i\in G_K$ are arbitrary elements where $\gamma\in \Gamma$ and $i\in\{1,\dots, m\}$. The result of the action $\gamma \circ g_i(x)$ lies in $H'$ because for each $k=1,\dots, m$ we have $g_k^{-1}\gamma g_i=\gamma'g_j^{-1}$ for some $\gamma'\in \Gamma$ and $j\in\{1,\dots, m\}$ so $\gamma g_i(x)\in \gamma g_ig_j(H)=g_k\gamma'(H)=g_k(H)$.

We can think of this argument as of a generalization of the proof that a compact group acting on a finite-dimensional $\mathbb{Q}_p$-vector space always preserves some $\mathbb{Z}_p$-lattice.