Let $G$ be a (connected) reductive linear algebraic group over $\overline{\mathbb{Q}}_p$. By definition, this means that $G$ is a closed subgroup of some $\mathrm{GL}_n$. We can always find a reductive $\mathbb{Z}_p$-group $\mathbb{G}$, which after base-change to $\overline{\mathbb{Q}}_p$ recovers $G$. I.e., $\mathbb{G}_{\overline{\mathbb{Q}}_p} = G$.

Let $K/\mathbb{Q}_p$ be an extension (inside $\overline{\mathbb{Q}}_p$) with ring of integers $R$. We have an inclusion $$ \mathbb{G}(R) \subset \mathbb{G}(\overline{\mathbb{Q}}_p) = G(\overline{\mathbb{Q}}_p)\subset \mathrm{GL}_n(\overline{\mathbb{Q}}_p). $$ Is there any relationship between $\mathbb{G}(R)$ and the intersection $\mathbb{G}(\overline{\mathbb{Q}}_p)\:\cap\:\mathrm{GL}_{n}(R)$ inside $\mathrm{GL}_n(\overline{\mathbb{Q}}_p)$? In particular, is the latter contained in the former? If something like this isn't true in general, is it true for a particular choice of $\mathbb{Z}_p$-model $\mathbb{G}$? Is there anything I can do to achieve a similar inclusion? Any relevant reference is greatly appreciated.