Yes. Take $f\in O_K$ a uniformizing element of some prime $\mathfrak p$. Consider the hyperelliptic curve defined by the equation
$$y^2=x^{2g+1}+f.$$
Then this curve doesn't have semi-stable reduction at $\mathfrak p$. In fact, this equation defines a proper regular model of the curve over the localization $O_{K, \mathfrak p}$ and this model is minimal because its closed fiber is irreducible (defined by $y^2=x^{2g+1}$), and it is not semi-stable. If the curve had semi-stable reduction, the minimal regular model would also be semi-stable. Further, this curve is far from being semi-stable because its Jacobian has purely additive reduction at $\mathfrak p$.

**Add** a more elementary explanation on why the curve doesn't have semi-stable reduction at $\mathfrak p$. Suppose for simplicity that $\mathfrak p$ is prime to $2(2g+1)$. As saw in the comments, the curve has potentially good reduction above $\mathfrak p$. So if it had already semi-stable reduction over $K$, then it would already have good reduction over $K$. This implies (as for elliptic curves) that after a suitable homographic transformation on $x$, we will get a new polynomial with discriminant invertible in $O_{K, \mathfrak p}$. But such transformation changes the valuation of the discriminant by a multiple of $2(2g+1)$ while the initial discrimiant has valuation $2g$. Impossible.