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.



Generalization of a lemma of Livne
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.


