Probably the easiest case to understand is when $\mathbf{G}$ is an inner form that is adjoint. In this case, there will always be infinitely many orbits, unless $\mathbf{G}$ manages to be simply connected (even though it is adjoint), which means that every simple factor of $\mathbf{G}$ is of type $E_8$, $F_4$, or $G_2$. (For a different case, see later in this answer for a discussion of all split groups.)

We are given a reductive group $\mathbf{G}$ over $\mathbb{Q}_p$, and a faithful representation $\rho \colon \mathbf{G} \to \mathbf{GL}_n$ that is defined over $\mathbb{Q}_p$. Assume, for now, that $\mathbf{G}$ is both inner and adjoint.

Suppose $\mathbf{H}$ is any inner $\mathbb{Q}$-form of $\mathbf{G}$, so we can think of $\rho$ as a representation of $\mathbf{H}$ that is defined over $\mathbb{Q}_p$. Since $\mathbf{H}$ is inner and adjoint, every representation of $\mathbf{H}$ can be realized over $\mathbb{Q}$. (A classic paper of Tits [J. Reine Angew. Math. 247 (1971) 196-220] describes the two obstructions to being able to realize a representation over the algebraic closure as a representation over the ground field. One obstruction comes from the $*$-action, which is trivial for inner forms, and the other comes from the center, which is trivial for adjoint groups.) Therefore, from the other answer, which relates orbits to $\mathbb{Q}$-forms, we just need to show that $\mathbf{G}$ has infinitely many different $\mathbb{Q}$-forms that are inner.

For each $\ell$ in a finite set $S$ of primes, choose an algebraic group $\mathbf{G}_\ell$ over $\mathbb{Q}_p$ that is isomorphic to $\mathbf{G}$ over an algebraic closure (and is inner). The proposition on page 525 of [Borel-Harder, J. Reine Angew. Math. 298 (1978) 53-64] implies that there is an inner $\mathbb{Q}$-form of $\mathbf{G}$ that is $\mathbb{Q}_\ell$-isomorphic to $\mathbf{G}_\ell$ for every $\ell$ in $S$. Assuming there is a simple factor of $\mathbf{G}$ that is not of type $E_8$, $F_4$, or $G_2$, then there is more than one possible choice of $\mathbf{G}_\ell$, for each prime $\ell$. Hence, there must be infinitely many different $\mathbb{Q}$-forms of $\mathbf{G}$ that are inner.

Proposition. Suppose that $\mathbf{G}$ is split (over $\mathbb{Q}_p$) and simply connected, and that $\rho$ is irreducible (and almost faithful). Let $\lambda$ be a highest weight of $\rho$. Then $\mathbf{G}(\mathbb{Q}_p) \times \mathbb{Q}^\times$ has finitely many orbits on the set of $\mathbb{Q}$-lattices for $\mathbf{G}$ if and only if $\rho$ is faithful and $\lambda$ is not fixed by any nontrivial diagram automorphism.

Proof. Since $\mathbf{G}$ is split, we know that $\rho$ is absolutely irreducible, not just irreducible, so $\mathbf{C}_\rho = \mathbb{Q}_p^\times$. (Also, the highest weight $\lambda$ is unique.) Therefore, the other answer shows that $\mathbf{G}(\mathbb{Q}_p) \times \mathbb{Q}^\times$ has finitely many orbits if and only if there are only finitely many $\mathbb{Q}$-forms of $\mathbf{G}$ such that $\rho$ can be realized as a representation over $\mathbb{Q}$.

Let $\mathbf{H}_0$ be the split $\mathbb{Q}$-form of $\mathbf{G}$.

($\Leftarrow$) Suppose $\mathbf{H}$ is $\mathbb{Q}$-form of $\mathbf{G}$ such that $\rho$ can be realized as a representation over $\mathbb{Q}$. Then Lemma 7.4 of the Tits paper tells us that $\lambda$ must be fixed by the $*$-action corresponding to $\mathbf{H}$. However, by assumption, $\lambda$ is not fixed by any nontrivial diagram automorphisms, so this implies that the $*$-action is trivial, which means that $\mathbf{H}$ is an inner form.

Therefore, $\mathbf{H} = {}^\zeta \mathbf{H}_0$ is the Galois twist of $\mathbf{H}_0$ by some cocycle $\zeta \in H^1(\mathbb{Q}; \overline{\mathbf{H}_0})$, where $\overline{\mathbf{H}_0}$ is the adjoint group. Letting $Z$ be the center of $\mathbf{H}_0$, we obtain a cohomology class $c \in H^2(\mathbb{Q}; Z)$. Composing with $\lambda$ yields a cohomology class $\lambda \circ c \in H^2(\mathbb{Q}; \mu)$, where $\mu$ is the group of all $n$th roots of unity in $\overline{\mathbb{Q}}$.

Since the $*$-action is trivial, Corollary 3.5 of the paper of Tits states that $\rho$ can be realized over $\mathbb{Q}$ if and only if $\lambda \circ c$ is trivial (in the cohomology group $H^2(\mathbb{Q}; \mu)$). (This is one of the main results of the paper.)

Since $\lambda$ is faithful, and $\lambda \circ c$ is trivial, we conclude that $c$ is trivial (in the cohomology group $H^2(\mathbb{Q}; Z)$). This means that $\zeta$ lifts to a well-defined $1$-cocycle $\eta \in H^1(\mathbb{Q}; \mathbf{H}_0)$.

Since $\mathbf{H}_0$ is simply connected (because $\mathbf{G}$ is simply connected), we know that $H^1(\mathbb{Q}_\ell ; \mathbf{H}_0) = 0$ for every prime $\ell$ [PR, Thm. 6.4,, p. 284. ([PR] = Platonov-Rapinchuk's book *Algebraic Groups and Number Theory*) Therefore, the natural map $H^1(\mathbb{Q} ; \mathbf{H}_0) \to H^1(\mathbb{R} ; \mathbf{H}_0)$ is finite-to-one [PR, Thm. 6.15, p.~316]. Since $H^1(\mathbb{R} ; \mathbf{H}_0)$ is finite [PR, Thm. 6.14, p.~316], we conclude that $H^1(\mathbb{Q} ; \mathbf{H}_0)$ is finite. Hence, there are only finitely many possibilities for $\eta$. So there are only finitely many possibilities for $\zeta$, which means there are only finitely many possibilities for the $\mathbb{Q}$-form $\mathbf{H} = {}^\zeta \mathbf{H}_0$.

($\Rightarrow$) Suppose $\lambda$ is fixed by the outer automorphism corresponding to some nontrivial automorphism $\varphi$ of the Dynkin diagram. Since $\varphi$ has order 2 or 3, I think it is easy to see that there are infinitely many different homomorphisms $\zeta$ from $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}}/\mathbb{Q})$ onto $\langle \varphi \rangle$, such that $\zeta$ is trivial on $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$. (For example, this is easier than Lemma 1.9 of [Borel-Harder].) Each $\zeta$ represents a different cohomology class in $H^1(\mathbb{Q}; \mathop{\mathrm{Aut}} \mathbf{H}_0)$, and therefore represents a different quasi-split $\mathbb{Q}$-form ${}^\zeta H_0$ of $\mathbf{G}$. (We are using the fact that $\zeta$ is trivial on $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ to know that ${}^\zeta H_0$ is isomorphic to $\mathbf{G}$ over $\mathbb{Q}_p$.)
By construction, the $*$-action for ${}^\zeta H_0$ is given by $\varphi$, so $\lambda$ is fixed by the $*$-action. Since ${}^\zeta H_0$ is quasi-split, Tits gives no additional obstruction, so the representation $\rho$ with highest weight $\lambda$ can be realized over $\mathbb{Q}$ with respect to each of the infinitely many $\mathbb{Q}$-forms ${}^\zeta H_0$

Suppose $\lambda$ is not faithful, so the kernel is some nontrivial subgroup $Z'$ of $Z = Z(\mathbf{H}_0)$. Then infinitely many cohomology classes in $H^1(\mathbb{Q}; \mathbf{H}_0/Z')$ have trivial restriction to $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$. (Indeed, since $H^1(\mathbb{Q}_\ell; \mathbf{H}_0/Z')$ is nontrivial for every prime $\ell$ [PR, Thm. 6.20, p. 326], this follows from the surjectivity result in Theorem 1.7 of [Borel-Harder].) Hence, twisting by elements $\zeta$ of $H^1(\mathbb{Q}; \mathbf{H}_0/Z')$ yields infinitely many different $\mathbb{Q}$-forms ${}^\zeta \mathbf{H}_0$ of $\mathbf{G}$. The corresponding element~$c$ of $H^2(\mathbb{Q}; Z)$ belongs to $H^2(\mathbb{Q}; Z')$. Since $Z' = ker \lambda$, then $\lambda \circ c$ is obviously trivial. So Tits tells us that $\rho$ can be realized over $\mathbb{Q}$ with respect to each of these infinitely many $\mathbb{Q}$-forms.

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