Let $G/\mathbb{Z}_p$ be a connected reductive group, and let $T_1, \cdots, T_n \subset G_{\mathbb{Q}_p}$ be maximal tori. Suppose that we have precisely one maximal torus in each $G(\mathbb{Q}_p)$-conjugacy class of maximal tori; is it then true that $T_1(\mathbb{Z}_p), \cdots, T_n(\mathbb{Z}_p)$ (topologically) generate $G(\mathbb{Z}_p)$? Here by $T(\mathbb{Z}_p)$ I mean the $\mathbb{Z}_p$ points of the connected component of the identity of the Neron model $\mathcal{T}$ of $T$. As LSpice pointed out in the comments, we need the inclusions $\mathcal{T}_i \subset G$ to be defined over $\mathbb{Z}_p$.

If $G=\operatorname{GL_n}$, then this is true. Our conjugacy classes of maximal tori correspond to commutative semi-simple algebras $C/\mathbb{Q}_p$ of degree $n$ and $M_n(\mathbb{Z}_p)$ is (topologically) generated by $\{\mathcal{O}_C\}_{C \in \mathcal{C}}$, where $\mathcal{C}$ is the set of isomorphism classes of such $C$ (and we make some choice of $\mathcal{O}_C \to M_n(\mathbb{Z}_p)$ for each class). An argument using the $p$-adic exponential map now shows that the (topological) closure $H$ of the group generated by our tori contains $1+p \cdot M_n(\mathbb{Z}_p)$, and then all we have to do is show that $H$ surjects onto $\operatorname{GL}_n(\mathbb{F}_p)$.

More generally, I wonder if there is an analogous statement with $\mathcal{G}/\mathbb{Z}_p$ a parahoric group scheme.

alltori) condition—maybe something like (as investigated by Fintzen) $p \nmid \lvert W\rvert$, or something? $\endgroup$ – LSpice Aug 3 '20 at 21:03knowthat we would need all tori to split over tamely ramified extensions; I just guessed, and may be wildly off base. My point was that I didn't think a random $\mathbb Q_p$-torus could be given a $\mathbb Z_p$-model, but you're probably right—the connected Néron model should do it, I guess. (I'm not really comfortable, despite @BCnrd's heroic efforts ("Reductive group schemes"), with what 'connected reductive group', or even 'torus', means over $\mathbb Z_p$.) $\endgroup$ – LSpice Aug 4 '20 at 10:05