Let $A$ be any supersingular abelian variety over $\mathbb{F}_p$ and consider $B=A \otimes_{Z} \mathcal{O}_E$. Another way of saying this is that we are choosing a $\mathbb{Z}$-basis of $\mathcal{O}_E$ giving an injection $\mathcal{O}_E \to M_{2 \times 2}(\mathbb{Z})$, and we let $\mathcal{O}_E$ act on $A \times A$ via the natural action of M_{2 \times 2}(\mathbb{Z})$ on the product.

I think your guess for $J$ is right, and also that considering $G$ as a group scheme instead of an abstract group makes everything much more complicated. Since your group is generated by a single element $g$, the fixed points $X^G$ can be identified with the intersection of the diagonal $\Delta$ with the graph $\Gamma_g$ of $g$ inside $X \times X$.

@user491858 Thanks! My student has been able to check that the $p$-part of the class group is trivial in this case.

On the other hand, if you fix the lattice $\Lambda$, then in general the groups $\mathbb{G}(\mathbb{Z}_p)$ and $\mathbb{G}(\mathbb{Q}_p) \cap \operatorname{GL}(\Lambda)(\mathbb{Z}_p)$ have nothing to do with each other. For instance, there does not have to be a containment in either direction.

I was trying to say that we likely need to assume that our map is defined over $\mathbb{Q}_p$ in order for your question to have a reasonable answer. Given this, there should be some choice of lattice $\Lambda$ such that $\mathbb{G}(R)$ is equal to $\operatorname{GL}(\Lambda)(R) \cap \mathbb{G}(K)$ for all finite extensions $K$ with ring of integers $R$. For $K=\mathbb{Q}_p$ atleast, the existence of the lattice follows from via a standard argument from the compactness of $\mathbb{G}(\mathbb{Z}_p)$ (any compact group acting on a finite dimensional $\mathbb{Q}_p$-vector space fixes a lattice).

Let $V$ be an $n$-dimensional vector space over $\mathbb{Q}_p$. Assume that that the closed immersion $G \to \operatorname{GL}(V)_{\overline{\mathbb{Q}}_p}$ comes from a map $i:\mathbb{G}_{\mathbb{Q}_p} \to \operatorname{GL}(V)$. Then there should be a $\mathbb{Z}_p$-lattice $\Lambda \subset V$ such that $i$ extends (uniquely) to a closed immersion $\mathbb{G} \to \operatorname{GL}(\Lambda)$ over $\mathbb{Z}_p$; this is certainly not true for any lattice.

For genus $2$ curves there is the recent work of Boxer--Calegari--Gee--Pilloni, see arxiv.org/abs/1812.09269.

The paper "On Conjugacy Classes of Maximal Tori in Classical Groups" by Kazutoshi Kariyama (J. Algebra 125(1989), no.1, 133–149) is a very relevant reference. I think one can construct a counterexample in $\operatorname{SL}_n$ or $\operatorname{Sp}_{2n}$ from their main results by doing some class field theory.

Is it intentional that your number of variables $T_1, \cdots, T_d$ is the same as the number of power series $(F_1, \cdots, F_d)$?

I think $E$ should be the classifying stack of the (affine) group scheme $\underline{\mathbb{Z}_{\ell}}^{\times}:=\varprojlim_{n} (\mathbb{Z}/\ell^{n} \mathbb{Z})^{\times}$ in the pro-etale topology. To be precise, I think it should classify pro-etale $\mathbb{Z}_{\ell}$-torsors.

The stack given by $U \mapsto \{\text{groupoid of } \rho:G \to \operatorname{SL}_2(U)\}$ is an algebraic stack, which is the (stack) quotient of $\operatorname{Hom}(G,\operatorname{SL_2})$ by the natural conjugation action of $\operatorname{SL}_2$. If the order of $G$ is invertible in $K$, then this is a disjoint union of classifying stacks, corresponding to the stabilisers of the $\operatorname{SL}_2$-orbits in $\operatorname{Hom}(G,\operatorname{SL_2})$. See Appendix A1 and A2 of arxiv.org/pdf/2009.06708.pdf for more details.

... And the set of orbits is in bijection with the set of isomorphism classes of $\rho:G \to \operatorname{SL}_2(\overline{K})$ over an algebraic closure of $K$. Now consider the finite etale $K$-scheme $\pi_0(\operatorname{Hom}(G,\operatorname{SL_2}))$. Under the assumption that the order of $G$ is invertible in $K$, this should represent the (etale) sheafification(!) of the presheaf $U \mapsto \{\text{isomorphism classes of } \rho:G \to \operatorname{SL}_2(U)\}$. I suspect something similar should be true in the case of rigid analytic spaces.

In the world of schemes over a field $K$ the space $\operatorname{Hom}(G,\operatorname{SL_2})$ of homomorphisms $\rho:G \to \operatorname{SL_2}$ is representable by a scheme of finite type. [Choose generators $x_1, \cdots, x_n$, then $\operatorname{Hom}(G,\operatorname{SL_2})$ can be realised as a closed subspace of $\operatorname{SL_2}^n$ cut out by the relations.] If the order of $G$ is invertible in $K$, then this is in fact a disjoint union of (open and closed) $\operatorname{SL_2}$-orbits...

If $G$ is simple over $\mathbb{C}$ it suffices to take proper subsets of the affine Dynkin diagram. But for example if $G=G_1 \times G_2$ then we get a decomposition $\Delta=\Delta_1 \coprod \Delta_2$ of the affine Dynkin diagrams and the only subsets of $\Delta$ that are allowed are of the form $I_1 \coprod I_2$ where $I_j \subset \Delta_j$ is a proper subset. The general case has a similar form.