The starting point of this question is the (presumably) well-known theorem (the proof I know is from Abelian $\ell$-adic representations and elliptic curves from J-P.Serre in which it is a lemma for $n=2$ and an exercise for $n>2$; which suggests that the result was already classical in the 60s).

Theorem I: If $G$ is a subgroup of $\operatorname{SL}_{n}(\mathbb Z_{p})$ which surjects on $\operatorname{SL}_{n}(\mathbb F_{p})$ and if $p≥5$, then $G=\operatorname{SL}_{n}(\mathbb Z_{p})$.

The theorem is optimal with respect to all hypotheses in the sense that there exists proper subgroups of $\operatorname{SL}_{2}(\mathbb Z_{p})$ mapping onto $\operatorname{SL}_{2}(\mathbb F_{p})$ when $p=2,3$ (a fact that played a role in the original proof of the modularity of semi-stable elliptic curves by A.Wiles, if I am not mistaken) and in the sense that for all $n≥2$ and all prime $p$, there exist discrete valuation rings $A$ of mixed characteristic $(0,p)$ such that $\operatorname{SL}_{n}(A)$ contains proper subgroups mapping onto $\operatorname{SL}_{n}(A/\mathfrak m)$ (just take a completely ramified $A$ over $\mathbb Z_{p}$ and consider $\operatorname{SL}_{2}(\mathbb Z_{p})$ inside $\operatorname{SL}_{2}(A)$).

A slightly less well-known fact is that the theorem admits the following generalization, due to N.Boston.

Theorem II: Let $A$ be a complete local noetherian ring with finite residual characteristic $p\neq 2$. If $G$ is a closed subgroup of $\operatorname{SL}_{n}(A)$ which surjects on $\operatorname{SL}_{n}(A/\mathfrak m^2)$, then $G=\operatorname{SL}_{n}(A)$.

Again, this theorem is optimal in the sense that there exists a proper subgroup of $\operatorname{SL}_{2}(\mathbb Z_{2})$ surjecting on $\operatorname{SL}_{2}(\mathbb Z/4\mathbb Z)$.

Now my actual question. Let $A$ be a complete local noetherian ring of mixed characteristic $(0,p)$ with $p≠2$. Suppose $G$ is a closed subgroup of $\operatorname{SL}_{n}(A)$ which surjects on $\operatorname{SL}_{n}(A/\mathfrak m)$.

Among the pre-images in $G$ of non-identity unipotents elements in $\operatorname{SL}_{n}(A/\mathfrak m)$, is it true that there exists a unipotent element?

Granted theorem II, this is obviously true if $G$ maps onto $\operatorname{SL}_{n}(A/\mathfrak m^2)$. Without this hypothesis, it looks dubious to me but nevertheless, the obvious counterexamples to theorem I coming from ramified rings do not provide counterexamples to this claim and I confess that I don't quite know how to construct other counter-examples. Taking this into account, another more general point of view on the question would be the following.

What are the subgroups of $\operatorname{SL}_{n}(A)$ which do not map onto $\operatorname{SL}_{n}(A/\mathfrak m^2)$ but which do map onto $\operatorname{SL}_{n}(A/\mathfrak m)$?

Perhaps group cohomology of $\operatorname{SL}_{n}$ would help then. Any positive result, even in the case $n=2$ and $p>3$ would already be of interest to me.