Vaguely, an arithmetic surface is an algebraic curve with coefficients in some "ring of integers".
More formally it is an integral normal scheme $X$ of dimension 2 of finite type over a Dedekind domain $R$ with some additional properties over its reduction modulo the primes of $R$ (this reductions being the same as the fibers of the morphism $X\rightarrow \text{Spec } R$). You can see the actual definition at the Wikipedia page on arithmetic surfaces. As this question is local I think there is no harm in assume $R$ a DVR.
Also, if I understand correctly, given a variety $V$ and a point $p\in V$, the completion $\widehat{\mathcal{O}}_{V,p}$ of the local ring at $p$ is a way to understand the singularity of the variety at $p$. For example when the point $p$ is smooth this completion is isomorphic to $k[[x_1,\dots,x_n]]$ for $n=\dim V$ as was already discussed here.
Now take $x\in X$ a closed point in the arithmetic surface. My idea would be that $\widehat{\mathcal{O}}_{X,x}$ is something that is determined on the regularity of $x$ both as a point in $X$ and as a point in its special fiber $X_s$ (where $s$ is the closed point of $\text{Spec }R$). So my idea would be that, if we fix the type of regularity of $x$ at $X$ and at $X_s$ then $\widehat{\mathcal{O}}_{X,x}$ should also be fixed.
Thinking in this way I want to know if the following is true:
Let $R$ be a DVR with algebraically closed residue field $k$ and $X/R$ be an algebraic surface with semistable reduction. Then for all point $x\in X_s$ regular in $X$ we have either $$\widehat{\mathcal{O}}_{X,x}\cong R[[T]] \ \ \text{ or } \ \ \widehat{\mathcal{O}}_{X,x}\cong R[[S,T]]/(ST-\pi)$$ for an uniformizer $\pi$ of $R$.
Here semistable reduction means that the special fiber is reduced and the only singularities of it are nodes. The points with local completion $R[[T]]$ should be the ones with smooth reduction and the ones with local completion $R[[S,T]]/(ST-\pi)$ should be the nodes.
Edit: Added the hypothesis of algebraically closed as discussed in the comments.
