One can show the following:

**Theorem:** If the group given by the subpresentation is hyperlinear (in particular if it sofic), then it must be trivial.

The argument is based on an old Theorem of Gerstenhaber-Rothaus. I can only give a sketch here: Denote the group given by the subpresentation by $G$. We want to show that $G$ embeds into the group given by the full presentation to get a contradiction. Now, as $G$ is hyperlinear, one can satisfy the group multiplication on a finite subset of $G$ by unitaries up to any desired precision in the normalized Hilbert-Schmidt norm. Equivalently, $G$ embeds into a metric ultraproduct of unitary groups. Now, if unitaries $g_{m+1},\dots,g_n$ can be found which satisfy the relations $r_{m+1},\dots,r_n$, then the embedding of $G$ can be extended to homomorphism into the ultraproduct defined on $g_1,\dots,g_n$. In particular, $G$ embeds into the group given by the full presentation. Now, the fact that $g_{m+1},\dots,g_n$ can be found relies on the solvability of the equations $r_{m+1},\dots,r_n$ and can be proved along the lines of Gerstemhaber-Rothaus using degree theory and Hopf's computation of the cohomology of unitary groups.

Gerstenhaber-Rothaus proved some generalized form of the Kervaire-Laudenbach Conjecture for residually finite groups in

M. Gerstenhaber and O.S. Rothaus, The solution of sets of equations in groups, Proc. Nat. Acad. Sci. U.S.A. 48 (1962), 1531–1533.

The above argument is essentially only the observation that their argument extends without any problems to hyperlinear groups.