Let $R$ be a Noetherian semi local ring, and $I=\text{N}(R)$, $J=\text{Jac}(R)$. If $R/I$ is complete w.r.t. $J$-adic filtration then we have to show that $R$ is complete in $J$-adic filtration.
Thus we have to show that the natural map $f:R \to \varprojlim R/J^{n}$ is an isomorphism while we are given that $g:R \to \varprojlim R/(I+J^{n})$ is an isomorphism. By Artin-Rees lemma we already have $\bigcap_{n \geq 1}J^{n}=(0)$, hence $f$ is injective. It remains to show that $f$ is onto. Lets consider a coherent sequence $(x_n) \in \varprojlim R/J^{n}$ so that $x_{n+1}-x_n \in J^{n}$ which implies that $x_{n+1}-x_n \in I+J^{n}$, i.e., $(x_n)$ is an element in $\varprojlim R/(I+J^{n})$. Since $g$ is onto there is an element $x \in R$ such that $x-x_n \in I+J^{n}$ for all $n \geq 1$. From here I can't complete the proof. I couldn't use that $J$ is intersection of only finitely many maximal ideals and $I \subset J$; also the fact $\bigcap_{n \geq 1}(I+J^{n})=I$ due to the map $g$ is injective. I need some help. Thanks.
The semi-local assumption is perhaps misleading because it is totally superfluous.
Much more generally, the following is true:
For a complete reference, see Stacks, 10.96.10.
For Noetherian rings, the nilradical is nilpotent. Any module is complete with respect to any nilpotent ideal, so in particular the cited fact implies that for Noetherian rings, completeness always lifts modulo the nilradical.