Let $A$ be a Noetherian commutative ring and $I$ an ideal in $A$. It is pretty much trivial to see that every free $A/I$-Module is obtained from a free $A$-module by tensoring over $A$ with $A/I$: Just choose preimages for every generator and take the free module generated by them. I'm wondering whether the same remains true if I replace ‘free’ by ‘projective and finitely generated’.

One way to prove this would be the following: Let $X=\operatorname{Spec} A$, $Z=\operatorname{Spec}(A/I)$, and $U=Z^c$. If there were something like an exact sequence

$$K_0(U) \rightarrow K_0(X) \rightarrow K_0(Z) \rightarrow 0$$

the claim would follow, right? Does such a sequence exist? If I'm not mistaken, such a sequence exists if $A$ is a Dedekind ring. Just to make it clear, here is the precise question:

Question: Given a Noetherian commutative ring $A$, an ideal $I$ and a projective, finitely generated $A/I$-module $M$, does there exist a projective, finitely generated $A$-module $N$ such that $N/IN=M$?