Let $K$ be a finite extension of the $p$-adic numbers with valuation ring $\mathcal{R}$ and uniformizer $\pi$. Consider a smooth and connected rigid $K$-variety $X=Sp(A)$ and assume that the affine formal model $\mathfrak{X}=Spf(A^{\circ})$ is normal (i.e. $A^{\circ}$ is a normal integral domain). My question is whether the fact that $X$ admits a normal affine formal model is stable under finite extensions of the ground field. That is, let $L$ be a finite extension of $K$ (separable as $K$ is of characteristic zero) with valuation ring $\mathcal{R}_{L}$ and uniformizer $\omega$. Is it true that $X_{L}=Sp(A\otimes_{K}L)$ admits a normal formal model? Letting $A_{L}=A\otimes_{K}L$, is $A_{L}^{\circ}$ a normal ring in general? Under which conditions does it hold that $A^{\circ}_{L}= A^{\circ}\otimes_{\mathcal{R}}\mathcal{R}_{L}$?
Context: By the reduced fiber theorem there is a finite extension $L$ of $K$ such that $A^{\circ}_{L}$ has geometrically reduced special fiber. As this process introduces roots of $\pi$, the induced morphism $\mathcal{R}\rightarrow \mathcal{R}_{L}$ will not be étale, just finite flat. Thus, the morphism $A^{\circ}\rightarrow A^{\circ}\otimes_{\mathcal{R}}\mathcal{R}_{L}$ is not neccesarily étale, and $A^{\circ}\otimes_{\mathcal{R}}\mathcal{R}_{L}$ is not necessarily normal. I am interesting in finding out if for every smooth irreducible rigid $K$-variety with a normal affine formal model with geometrically irreducible special fiber there is a finite extension $L$ such that $X_{L}$ admits an affine formal model with integral geometric fiber.