**EDIT** I may have overlooked the assumption that $L$ is contained in $\hat{K}$. In general, if $v$ is a valuation of $K$ and $L/K$ is finite separable then $L \otimes_K \hat{K}$ is reduced and thus isomorphic to $\prod_i \hat{L}_i$ where the $\hat{L}_i$ are the completions of $L$ for the various valuations $v_i$ extending $v$. In your situation one of the $\hat{L}_i$ is equal to $\hat{K}$. Since the dimension of $L \otimes_K \hat{K}$ over $\hat{K}$ is equal to $[L:K]$, this means that there must exist some other completions, hence the extension of $v$ is not unique in your setting and the following examples don't apply directly. I'm leaving them however because they might be useful to answer the question. **END OF EDIT**

The following example, due to Ostrowski and mentioned in *The theory of classical valuations* by Ribenboim (Sect. 6.3), gives a situation where $V$ is not finitely presented over $R$, and thus not étale over $R$.

Let $K=\mathbb{Q}_2(2^{1/2^\infty})$ be the extension of $\mathbb{Q}_2$ obtained by adjoining the iterate square roots of $2$, endowed with the $2$-adic valuation $v$, with value group $\mathbb{Z}[\frac12]$, so $v$ has rank 1 and is not discrete. Consider the finite separable extension $L=K(\sqrt{-1})$. Then $v$ has a unique extension $v'$ to $L$ with ramification index and residual degree both equal to 1. Let $R$ be the valuation ring of $v$. Since $v$ has a unique extension to $L$, the integral closure $V$ of $R$ in $L$ is the valuation ring of $v'$. We are in a situation where $\sum_i e_i f_i < [L:K]$ using standard notations from ramification theory, and this implies that $V$ is not a finite $R$-module (see Endler, *Valuation theory*, Theorem 18.6).

Just after Ostrowski's example, Ribenboim gives another interesting example of a complete valued field $(K,v)$ of characteristic $p$ with $v$ non-discrete, and a separable extension $L/K$ of degree $p$, such that $v$ has a unique extension $v'$ in $L$, and the ramification index and the residual degree of $v'/v$ are both equal to 1. This gives another example where $V$ is not étale over $R$.

Since in your situation finite presentation is the only obstruction, one may wonder whether pro-étale extensions are more appropriate, in particular whether $V$ is always pro-étale over $R$.

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