The answer is Yes. We denote $K(G)=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$. We compute $K(G)$; see the corollary below. It is clear from the corollary that $K(G)$ is canonically isomorphic to $K(H)$.
We will use Section 3 of M. Borovoi, Abelian Galois cohomology of reductive groups. Memoirs of the AMS 132 (1998), No. 626, although all necessary results can be found in Deligne's paper Variétés de Shimura: interprétation modulaire, et techniques de construction de modèles canoniques, Proc. Sympos. Pure Math. 33, Part 2, pp. 247–289.
We consider the crossed module $(G^{\rm sc}\to G)$ and the hypercohomology $$H^0_{\rm ab}({\mathbb Q},G):=H^0({\mathbb Q},G^{\rm sc}\to G),$$ where $G$ is in degree 0; see the Memoir. By definition $H^0_{\rm ab}({\mathbb Q},G)$ is a group. We consider the abelian crossed module $(Z^{\rm sc}\to Z)$, where $Z=Z(G)$ and $Z^{\rm sc}=Z(G^{\rm sc})$. The morphism of crossed modules $$(Z^{\rm sc}\to Z)\,\longrightarrow\,(G^{\rm sc}\to G)$$ is a quasi-isomorphism, and hence it induces a bijection on hypercohomology, permitting us to identify $H^0_{\rm ab}({\mathbb Q},G)$ with the abelian group $H^0({\mathbb Q},Z^{\rm sc}\to Z)$. We conclude that $H^0_{\rm ab}({\mathbb Q},G)$ is naturally an abelian group and that it does not change under inner twisting of $G$.
The short exact sequence $$1\to(1\to G)\to (G^{\rm sc}\to G)\to (G^{\rm sc}\to 1)\to 1$$ (where $G^{\rm sc}\to 1$ is not a crossed module) induces a hypercohomology exact sequence $$ G^{\rm sc}({\mathbb Q})\to G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc}),$$ where $${\rm ab}^0\colon G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)$$ is the abelianization map. This permits us to identify $G({\mathbb Q})/\rho G^{\rm sc}({\mathbb Q})$ with the kernel $${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$$ (yes, this kernel is a subgroup of the abelian group $H^0_{\rm ab}({\mathbb Q},G)$ ). This kernel might change under inner twisting of $G$, because $H^1({\mathbb Q},G^{\rm sc})$ changes under inner twisting.
By definition, $G({\mathbb R})_+=Z({\mathbb R})\cdot\rho G^{\rm sc}({\mathbb R})$, and hence $$G({\mathbb R})_+/\rho G^{\rm sc}({\mathbb R})={\rm ab}^0(Z({\mathbb R}))\subset {\rm ker}[ H^0_{\rm ab}({\mathbb R},G)\to H^1({\mathbb R}, G^{\rm sc})].$$ We see that $K(G):=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$ can be identified with the preimage of ${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$ in ${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$.
Lemma. The preimage of ${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$ in ${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$ coincides with the preimage of ${\rm ab}^0(Z({\mathbb R}))$ in $H^0_{\rm ab}({\mathbb Q},G)$.
Proof. Let $\xi\in H^0_{\rm ab}({\mathbb Q},G)$ lie in the preimage of $${\rm ab}^0(Z({\mathbb R}))\subset {\rm ker}[ H^0_{\rm ab}({\mathbb R},G) \to H^1({\mathbb R}, G^{\rm sc})].$$ Then the image of $\xi$ in $H^1({\mathbb R},G^{\rm sc})$ is trivial, and therefore, the image of $\xi$ in $H^1({\mathbb Q},G^{\rm sc})$ lies in the kernel of the localization map $$ H^1({\mathbb Q}, G^{\rm sc})\to H^1({\mathbb R},G^{\rm sc}).$$ By the Hasse principle for simply connected groups, this kernel is trivial. Thus the image of $\xi$ in $H^1({\mathbb Q},G^{\rm sc})$ is trivial, and hence $\xi$ lies in the preimage of ${\rm ab}^0(Z({\mathbb R}))$ in ${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$, as required.
Corollary. The abelianization map ${\rm ab}^0\colon G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)$ with kernel $\rho G^{\rm sc}({\mathbb Q})$ induces a canonical isomorphism between the abelian groups $K(G):=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$ and the preimage of ${\rm ab}^0(Z({\mathbb R}))$ in $H^0_{\rm ab}({\mathbb Q},G)$.