If $X$ has no (potentially) good reduction, then the answer to your question is no. More precisely, there always exists a finite cover $Y\to X$ such that for any finite extension $L/K$, no regular semi-stable model of $Y_L$ dominates a regular semi-stable model of $X_L$.

Suppose we are given a finite morphism of semi-stable models $\mathcal Y\to\mathcal X$ with $\mathcal Y$ regular but $\mathcal X$ is singular. I claim that in this case, no semi-stable regular model of $Y$ can dominate a semi-stable regular model of $X$ dominating $\mathcal X$.

For simplicity, I will work over a strictly henselian DVR. Let $x_0$ be a singular point of $\mathcal X$ and let $y_0$ be a point of $\mathcal Y$ lying over $x_0$. Then $x_0$ is a double point in its fiber $\mathcal X_s$ and similarly for $y_0$. Let $\mathcal Y'$ be a regular semi-stable model dominating $\mathcal Y$ and dominating a desingularization of $\mathcal X$. As $\mathcal X'\to \mathcal X$ is not an isomorphism above $x_0$, $\mathcal Y'\to\mathcal Y$ is not an isomorphism above $y_0$. So some irreducible component $\Gamma$ of
$\mathcal Y'_s$ must be contracted to $y_0$ in $\mathcal Y$. A smooth point $\mathcal Y'$ contained in $\Gamma$ lifts to a section of $\mathcal Y'$. This section is mapped to a section of $\mathcal Y$ passing through $y_0$. But as $\mathcal Y$ is regular, its sections are contained in its smooth locus. Contradiction because $y_0$ is not a smooth point.

With some extra work, one can show that the same property holds over any finite extension $L/K$ (one has to desingularize first the singular points of $\mathcal Y_{\mathcal O_L}$.).

Finally, for any $X$ with potentially bad reduction at a prime $\mathfrak p$ and for any integer $n\ge 2$ prime to $\mathfrak p$, there exists (after enlarging $K$) a cyclic étale cover $Y\to X$ of degree $n$ with the required property on the double points. See §6.3 and especially Prop. 6.6 in this paper.