$\newcommand{\nN}{{\mathcal N}}
\newcommand{\SL}{{\rm SL}}
\newcommand{\G}{{\bf G}}
\newcommand{\Q}{{\Bbb Q}}
\newcommand{\R}{{\Bbb R}}
\newcommand{\C}{{\Bbb C}}
$The answer is **No**.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$.
We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that 
$\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$.
Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$.
Since $\nN$ acts on $\G$ by *inner* automorphisms 
(see the comment of @YCor below),
we see that $\G'$ must be an anisotropic *inner form* of $\G$ 
with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where 
$D$ is a central simple algebra of degree 4 over $\Q$.
Since $\G'$ is anisotropic, $D$ is a division algebra.
However, then $\G'=\SL(1,D)$ has no 8-dimensional representation over $\Q$ containing over $\C$ the standard 4-dimensional representation of $\G'_\C=\SL_{4,\C}$.
Thus there is no anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.