The classification of forms of algebraic groups is generally only done for Galois extensions $L/K$. One gets a description of all $L/K$-forms of an algebraic $K$-group in terms of Galois cohomology.
But how do I find the algebraic $\mathbb{R}/\mathbb{Q}$-groups of some fixed algebraic $\mathbb{Q}$-group $\mathbf{G}$ ? I thought maybe it is enough to classify the $\overline{\mathbb{Q}}/\mathbb{Q}$-forms of $\mathbf{G}$, but then I would need something like this:
Let $\mathbf{G}$ and $\mathbf{H}$ be linear algebraic $\mathbb{Q}$-groups which are isomorphic as $\overline{\mathbb{Q}}$-groups. Then are they also isomorphic as $\mathbb{R}$-groups, or are at least $ \mathbf{G}(\mathbb{R})$ and $\mathbf{H}(\mathbb{R})$ isomorphic as real Lie groups?
If there is no general answer, how does one proceed in the case $\mathbf{G} = \mathbf{SL_n}$?
I think you didn't ask the question you meant to ask. Of course it's not true that $G_{\overline{\mathbb{Q}}} \cong H_{\overline{\mathbb{Q}}}$ implies $G_{\mathbb{R}} \cong H_{\mathbb{R}}$; a simple counterexample is $G = \mathbb{G}_m$, $H = \left\{ \begin{pmatrix} a & b \\ -b & a \end{pmatrix}: a^2 + b^2 = 1\right\}$ -- there is an isomorphism between the two of these over $\mathbb{Q}(i)$ given by sending $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ to $a + i b$, but $G(\mathbb{R})$ and $H(\mathbb{R})$ are trivially not isomorphic as real Lie groups, because one is compact and the other is not.
What you probably meant to ask is the converse: if $G_{\mathbb{R}} \cong H_{\mathbb{R}}$, does this imply $G_{\overline{\mathbb{Q}}} \cong H_{\overline{\mathbb{Q}}}$? This is rather less obvious, but it is still true. Intuitively, the reason it holds is this: to find an isomorphism $G_K \cong H_K$ for some field $K$, you have to find solutions in $K$ to some finite collection of polynomial equations (with coefficients in $\mathbb{Q}$); and if these have solutions over $\mathbb{R}$, they must have solutions over some finite extension of $\mathbb{Q}$.