The answer is affirmative in the "dominated by" aspect, with $R$ any complete discretely valued field whose fraction field $K$ has characteristic 0 and residue field $k$ has characteristic $p \ge 0$ (using $e$ not divisible by $p$; i.e., $e \in k^{\times}$). One cannot do better since the absence of various roots of unity from $K$ can mess up the "easy" Galois theory description of intermediate extensions. This is a standard application of Abhyankar's Lemma, as follows.
In the ring $A = R[\![x]\!]$, the height-1 prime $P = (x)$ has residue field $K$ of characteristic 0 (and uniformizer $x$), so all ramification at $P$ in any finite extension $L'$ of $L := {\rm{Frac}}(A) = {\rm{Frac}}(R(\!(x)\!))$ is tame. Hence, if $A'$ is the (module-finite) normalization of $A$ in $L'/L$ and $A'[1/x]$ is etale over $A[1/x] = R(\!(x)\!)$ then there is only tame ramification over ${\rm{Spec}}(A)$. It follows by Abhyankar's Lemma (as proved in an appendix of the book of Freitag-Kiehl on etale cohomology, for example) that ${A'}^{\rm{sh}} = A^{\rm{sh}}[x^{1/e}]$ for an integer $e \ge 1$ with $e \in k^{\times}$. Thus, $A' \subset A^{\rm{sh}}[x^{1/e}]$.
Since $A$ is complete local noetherian, so $A^{\rm{sh}}$ is exhausted by local finite etale extensions of $A$, which in turn correspond to finite separable extensions of $k$, as we vary through the local finite etale extensions $R'$ of $R$ the resulting finite etale $A$-algebras $R'[\![x]\!]$ exhaust $A^{\rm{sh}}$. Hence, for big enough such $R'$ we see that $A' \subset R'[\![x]\!][x^{1/e}]$.
The extension of Dedekind domains
$$R(\!(x)\!) = A[1/x] \rightarrow R'(\!(x^{1/e})\!)$$
is clearly finite etale, so any intermediate module-finite Dedekind domain is also finite etale over $A[1/x]$. Thus, the finite etale $R(\!(x)\!)$-algebra domains are precisely the module-finite subalgebras of such extensions $R'(\!(x^{1/e})\!)$ for $R'$ local finite etale over $R$ and $e \ge 1$ not divisible by ${\rm{char}}(k)$. That is the "dominated by" assertion we wanted to prove.
