$\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 a [comment](https://mathoverflow.net/questions/456984/mathbbq-forms-of-operatornamesl-4-mathbbr-inside-operatornamesl#comment1184403_456984) of @YCor above),
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, a representation $\rho$ of $\G'=\SL(1,D)$ over $\Q$ such that $\rho\otimes_\Q\C$ contains the standard 4-dimensional representation of $\G'_\C\simeq\SL_{4,\C}$ in $\C^4$, has dimension at least 16. 
Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.