Update: As mentioned below, the answer to the original question is a strong No. However, the case of $\pi_4$ remains, and actually I think that this one would follow from Suslin's conjecture on injectivity of $H_i(\mathrm{GL}_n(F))\to H_i(\mathrm{GL}_{n+1}(F))$ up to torsion (in the case $i=4$, $n=3$) together with some results of Dupont (specifically exact sequence (15.10) in Dupont's book "Scissors congruences, group homology and characteristic classes") and Goncharov's conjecture on motivic cohomology in terms of Bloch groups in weights $2$ and $3$. Concretely:
Question. Let $F$ be any field. Consider the inclusion $F^\times\to \mathrm{PGL}_2(F)$ of the diagonal matrices. Is the induced map $$ \bigwedge^4 F^\times_{\mathbb Q} = H_4(F^\times,\mathbb Q)\to H_4(\mathrm{PGL}_2(F),\mathbb Q)$$ surjective?
(Explanation: In Dupont's (15.10), the kernel of the last map is precisely $H^2(F,\mathbb Q(3))$ under Goncharov's conjecture. Under Suslin's conjecture, $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ is injective, and the target is stable. Computing in terms of the homology of the $K$-theory space, one sees that the contributions to $H_4(\mathrm{PGL}_4(F),\mathbb Q)$ are $\mathrm{Sym}^2 K_2^M(F)_{\mathbb Q}$ and $H^2(F,\mathbb Q(3))$ and $K_4^M(F)_{\mathbb Q}$. The last term $K_4^M(F)_{\mathbb Q}$ is precisely the cokernel of the map $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ by Suslin's theorem, so $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ should have contributions from $H^2(F,\mathbb Q(3))$ and $\mathrm{Sym}^2 K_2^M(F)_\mathbb{Q}$. The latter admits a surjection from $\mathrm{Sym}^2(\bigwedge^2 F^\times)_{\mathbb{Q}} = H_4((F^\times)^2\rtimes \Sigma_3)$, which is precisely the decomposable part of $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ in Dupont's work. Looking back at (15.10), this shows that the first term should be zero, which leads to the question. There is some confusion between $\mathrm{PSL}$ and $\mathrm{PGL}$ in what I said, but at least assuming that all elements of $F$ have $6$-th roots, the argument should work.)
Here is a hopelessly naive question. Please point me to the relevant literature!