I only vaguely know about étale fundamental groups at the moment so I went the other way and computed $H^1(X,\mathbb{Z})$. Let $\pi : Y=\mathrm{Spec}(\mathcal{O}_K) \to X$ denote the normalization. Denote $Z$ the singular locus and put $s_v=\#\pi^{-1}(v)-1$ for every $v\in Z$.

**Claim** : $H^1(X,\mathbb{Z})$ is a free abelian group of rank $\sum_{v\in Z}s_v$.

*Proof :* Consider the cokernel $F$ of the injection $0\to \mathbb{Z} \to \pi_\ast \mathbb{Z}$. Then $F$ is a skyscraper sheaf supported on singular points and we can compute by Galois cohomology that $H^0(X,F)\simeq \oplus_{v\in Z}\mathbb{Z}^{s_v}$ (this comes from the vanishing of the $H^1$ in Galois cohomology with coefficients in $\mathbb{Z}$). Because $H^1(Y,\mathbb{Z})=0$ since $Y$ is normal (see etale-cohomology-with-coefficients-in-the-integers), the long exact sequence in cohomology reads
$$
0 \to \mathbb{Z} \xrightarrow{\mathrm{Id}} \mathbb{Z} \to H^0(X,F) \to H^1(X,\mathbb{Z}) \to 0
$$ which proves the claim. $\blacksquare$

In particular, as soon as there is a singular point with at least two primes above it in the normalization, the (SGA3) étale fundamental group of $X$ be **not** profinite.