The answer is YES.

**Theorem 1.** Let $G$ be a connected semisimple linear algebraic group defined over a number field $k\subset{\mathbb{R}}$.
There exists a natural number $d=d(G_{\bar{k}})$ with the following property:
let $H_1$ and $H_2$ be two connected semisimple $k$-subgroups of $G$ that are conjugate over ${\mathbb{R}}$,
then there exists a finite real field extension $K=K(H_1,H_2)\subset {\mathbb{R}}$ of $k$ of degree $[K:k]\le d$ such that $H_1$ and $H_2$ are conjugate over $K$.

Since there are only finitely many conjugacy classes of connected semisimple subgroups in $G_{\bar{k}}$, see Friedrich Knop's answer to
this question,
Theorem 1 follows from the next Theorem 2 (as in my previous answer).

**Theorem 2.** Let $N$ be a linear algebraic group (not necessarily connected or reductive) over a real number field $k\subset{\mathbb{R}}$. The there exists $d=d(N_{\bar{k}})$
such that any cohomology class $\xi\in H^1(k,N)$ that vanishes over ${\mathbb{R}}$ can be killed by a finite real field extension $K=K(N,\xi)\subset{\mathbb{R}}$ of degree $[K:k]\le d$.

Note that the analog of Theorem 2 for abelian varieties is false.

We need three lemmas.

**Lemma 1.** Let $N$ be a *finite* $k$-group. Then any ${\mathbb{R}}$-point of $N$ is defined over a real number field of bounded degree.

This is obvious.

**Lemma 2.** If $N$ is a *connected* linear algebraic $k$-group, then $N(k)$ is dense in $N({\mathbb{R}})$.

See Sansuc's paper,
Cor. 3.5(iii).

*Lemma 3.* If $N$ is a *connected or abelian* linear algebraic $k$-group, then the localization map
$$ H^1(k,N)\to H^1({\mathbb{R}},N) $$
is surjective.

See Sansuc's paper, Lemma 1.6 and Lemma 1.12.

Note that the analog of Lemma 3 for nonabelian finite $k$-groups is an open question.

With the help of Lemmas 1, 2, and 3, we use dévissage in order to reduce Theorem 2 to the following four propositions.

**Proposition 1.** Theorem 2 is true when $N$ is a finite $k$-group.

Indeed, then any $k$-torsor $\mathcal{T}$ of $N$ is a finite $k$-scheme, hence any ${\mathbb{R}}$-point of $\mathcal{T}$ is real algebraic of bounded degree.

Now let $N$ be connected, then it splits over an algebraic extension $L\subset \mathbb{C}$ of bounded degree. Set $K=L\cap{\mathbb{R}}$.
After passing to $k=K$, we may assume that $N$ splits over $k$ or over a quadratic extension $L$ of $k$.
After passing to a real quadratic extension, we may assume that $k$ has only one real embedding.

**Proposition 2.** Theorem 2 is true when $N$ is a simply connected semisimple $k$-group.

Indeed, under our assumptions the localization map
$$ H^1(k,N)\to H^1({\mathbb{R}},N) $$
is bijective (the Hasse principle for simply connected groups), hence $\xi=1$.

**Proposition 3.** Theorem 2 is true when $N$ is a $k$-torus.

*Proof.* We assume that $N$ splits over a quadratic extension $L$ of $k$.
Then we may assume that $N$ is a nonsplit one-dimensional $k$-torus.
Write $L=k(\sqrt{-\alpha})$, where $\alpha\in k$, $\alpha>0$.
Then $N$ is given by the equation
$$ x^2+\alpha y^2=1,$$
and a $k$-torsor $\mathcal{T}$ of $N$ is given by the equation
$$ x^2+\alpha y^2=c,$$
where $c\in k^\times$.
By assumption $\mathcal{T}$ has an ${\mathbb{R}}$-point $(x_1,y_1)$,
and we may assume that $x_1,y_1>0$.
Choose $y_2\in k$ such that $0< y_2<y_1$ and set $\zeta=\sqrt{c-\alpha y_2^2}$,
then $K:=k(\zeta)$ is a real quadratic extension of $k$, and $\mathcal{T}$ has a $K$-point $(\zeta, y_2)$. This $\mathcal{T}$ splits over a real quadratic extension, as required.

**Proposition 4.** Theorem 2 is true for $H^2(k,A)$, where $A$ is a finite abelian $k$-group.

*Proof.* Set $A'=\mathrm{Hom}(A,\mathbb{G}_m)$, the dual group. We say that $A$ is *split* over $k$ if the Galois group $G_k$ acts trivially on $A'$.
As above, we may assume that $A$ splits over a quadratic extension $L$ of $k$ and that $k$ has only one real embedding.
Then, since $A$ splits over a quadratic extension, the Hasse principle for $H^1(k,A)$ and $H^2(k,A)$ holds.
The Galois group $G_k$ acts on $A'$ via its quotient group $\Gamma=G_{L/k}$ of order 2. We may assume that $A'$ is a *simple* $\Gamma$-module, i.e., is has no proper
$\Gamma$-submodules. Then $\#A=\ell$ is a prime number, and either $\Gamma$ acts on $A'$ trivially,
or $\ell$ is odd and $\Gamma$ acts on $A'$ by multiplication by $-1$.
The first case was treated in my previous answer: we can kill a cohomology class
$\eta\in H^2(k,A)$ that vanishes over $L$ by a real extension of degree $\ell$. Similarly, in the second case case one can kill a cohomology class $\eta$ that vanishes over $L$ by a real extension of degree $\ell$.
(One should use the Tate duality and the Tate-Poitou exact sequence, see Serre's book "Galois cohomology".)

This completes the proofs of Theorem 1 and Theorem 2.