After reading Will Sawin's comment, I think there may be an explanation in the spirit of what you seem to be looking for. The group $\operatorname{H}^1(G,R(L))$ classifies $L/K$-twists of the group scheme $R$. However, $R$ is itself an $L/K$-twist of $\mathbb{G}_m^n$, so the same group classifies $L/K$-twists of $\mathbb{G}_m^n$, and those are all trivial by generalized Hilbert 90 (i.e. by the fact that $\operatorname{H}^1(G,\operatorname{GL}_n(L))=0$). 
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**Previous answer:** There is a general argument that is slightly more elementary than what you wrote. By standard properties of Weil restrictions, we have $R(L) = \prod_{\sigma} \mathbb{G}_m(L)$, where the product is taken over the different $K$-linear embeddings of $L$ into some algebraic closure $\overline{K}$ and where $G$ acts on the factors in a natural way. In other words, we have that $R(L) = \mathbb{Z}[G] \otimes \mathbb{G}_m(L)$, as $G$-modules. By an easy exercise, we have that $$\mathbb{Z}[G] \otimes \mathbb{G}_m(L) = \operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L)),$$
giving $R(L)=\operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L))$. (Indeed, take the isomorphism
$$
f:\mathbb{Z}[G] \otimes \mathbb{G}_m(L) \rightarrow \mathbb{Z}[G] \otimes (\mathbb{G}_m(L))_0 \stackrel{\operatorname{def}}{=} \operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L)),
$$
where $\mathbb{G}_m(L)_0$ denotes the underlying abelian group of $\mathbb{G}_m(L)$, satisfying $f(g \otimes x) = g \otimes g^{-1}x$.) 

In conclusion, $R(L)=\operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L))$. Finally, by Shapiro's lemma, the $\operatorname{H}^1$ (and in fact, all higher cohomology) of this vanishes.