Let $H$ be a finite $2$-group. Let $N_{4}(H)$ be the subgroup generated by fourth powers. Let $H_{4}$ be the last term in the short exact sequence $1\rightarrow N_4(H) \rightarrow H \rightarrow H_{4} \rightarrow 1$. Suppose that every element of $H_{4}$ has a lift to an element of order $4$, $2$ or $1$ in $H$. Is it then true that $N_4(H)=1$? When $4$ is replaced by $2$ this is true, and used in Livne's generalization of Faltings-Serre.

The answer is yes. First, notice that if $\phi:G\rightarrow G'$ is an epimorphism of 2-groups, then $\phi(N_4(G)) = N_4(G')$. Let now $H$ be the group in your statement. Assume that $N_4(H)$ is nontrivial. Then it contains a maximal subgroup $N$ of index 2 which is normal in $G$ (this follows from considering the action of the quotient $H_4$, which is also a 2 group, on the $\mathbb{F}_2$-vector space $N_4(H) / N_2(N_4(H))$). By considering the group $H/N$ it is enough to prove that if $|N_4(H)|=2$ and every element of $H_4$ is liftable in the way you describe we get a contradiction. So $N_4(H)$ contains a unique nontrivial element $z$, which moreover has order 2 and is central in $H$. Since $N_4(H)$ is generated by all 4 powers in $H$, there is a $h\in H$ such that $h^4=z$. So $h$ has order 8. We know by assumption that $\bar{h}\in H_4$ has a lift of order 2 or 4. But the only other possible lift is $hz$ which also has an order 8, a contradiction.