Let us work over $K = \mathbf{C}((t))$ for simplicity. We say that a smooth proper scheme $X/K$ has *good reduction* if it extends to a smooth and proper algebraic space $\mathcal{X}/\mathcal{O}\_K$ where $\mathcal{O}\_K = \mathbf{C}[[t]]$, and that it has *potentially good reduction* if for some finite extension $K' = \mathbf{C}((t'))$ ($t' = t^{1/N}$), the base change $X\_{K'}$ has good reduction over $K'$. 


My question may sound a bit silly:



> 
> **Question.** Can you give an example of a (smooth, projective) rational surface $X/K$ which does not have potentially good reduction?
> 
> 
> 


On the one hand, the usual "homotopical" obstructions to good reduction vanish. The only nontrivial one I can think of is the action of the Galois group of $K$ on the Neron-Severi group ${\rm NS}(X\_{\bar K})$, but that one factors through the action of a finite quotient, and hence becomes trivial over some finite extension $K'$.


On the other, it is easy to think of a possible culprit. Take a sufficiently complicated zero-dimensional subscheme $Z\_0\subseteq \mathbf{P}^2\_{\mathbf{C}}$, and pick a formal curve 
$$ z\colon \operatorname{Spec} \mathbf{C}[[t]] \longrightarrow \operatorname{Hilb}(\mathbf{P}^2\_\mathbf{C}), \quad z(0) = [Z\_0]$$
through $Z\_0$ and such that $z(\eta)$ ($\eta$ being the generic point) corresponds to a smooth subscheme $Z\_\eta \subseteq \mathbf{P}^2\_K$. Take $X$ (resp. $\mathcal{X}$) be the blowup of $\mathbf{P}^2\_K$ (resp. $\mathbf{P}^2\_{\mathcal{O}\_K}$) along $Z\_\eta$ (resp. the subscheme $Z$ corresponding to $z$). Then $\mathcal{X}$ will not be smooth if $Z$ is complicated enough. Of course, this does not imply that $X$ does not have potentially good reduction, and I do not see how to check this.



Good reduction of rational surfaces
As Ulrich commented, this is true. In fact, if $\mathcal{O}\_K$ is a Dedekind domain with fraction field $K$ and $X$ is a rational surface over $K$, then there is a finite field extension $L/K$ such that $X\_L$ has a smooth *projective* model over the normalization $\mathcal{O}\_L$ of $\mathcal{O}\_K$ in $L$. 


I just wanted to seize this opportunity to make a couple of additional comments.


1) Suppose that $X$ is a smooth *split* cubic surface over $\mathbb{Q}$, so that $X$ is the blow-up of $\mathbb{P}^2\_{\mathbb{Q}}$ in six points in general position. Then, $X$ has a smooth projective model over $\mathbb{Z}$. However, the smooth cubic surface $X$ does not have a smooth projective model over $\mathbb{Z}$ which is a smooth cubic surface over $\mathbb{Z}$. To prove this, it is good to have a look at Scholl's observation that having a smooth proper model does not imply having a smooth proper model with relatively ample anti-canonical bundle. This is in Remark 4.6 of Scholl's paper [here](https://academic.oup.com/jlms/article-abstract/s2-32/1/31/979128). A variation of Scholl's observation is given in Lemma 4.9 of this [paper](https://arxiv.org/abs/1505.02249). The fact that there really are not such smooth cubic surfaces over $\mathbb{Z}$ is proven below. Let me make another remark on what it means (or should mean) to have "good reduction" as a rational surface.


2) The fact that every smooth rational surface potentially has a good model (in your sense) means that the stack of such surfaces satisfies the existence part of the valuative criterion of properness. The comment above shows that the (algebraic) stack of del Pezzo surfaces does not satisfy this property. 


3) Here is how you show that there are no smooth cubic surfaces $X$ over $\mathbb{Z}$. First, note that any smooth cubic surface $X$ over $\mathbb{Z}$ is split. This follows from the fact that Hilbert scheme of lines $Lines\_{X/\mathbb{Z}}$ is finite étale over $\mathbb{Z}$ and that $\mathbb{Z}$ is simply connected (Hermite). Then, it follows that $X$ is the blow-up of six $\mathbb{Z}$-points of $\mathbb{P}^2\_{\mathbb{Z}}$ in general position. The coordinates of such points would (after some appropriate rescaling) be solutions to the unit equation in $\mathbb{Z}$. As the unit equation has no solutions in $\mathbb{Z}$ it follows that there are no smooth cubic surfaces over $\mathbb{Z}$.


