I'd like to ask your opinion whether this answer is correct. If not, under which conditions the following lemma is true?
Lemma Let $p$ be a prime number and $B$ a noetherian torsion free $\mathbb{Z}_p$-algebra. Let $M$ be a finitely generated $B$-module such that $M/pM$ is a free $B/pB$-module and $M$ is torsion-free as $\mathbb{Z}_p$-module. Then $M$ is a free $B$-module.
Proof Let $e_1,\dots,e_n$ be a $B/pB$-basis of $M/pM$. Let $N=(b_1,\dots,b_n)$ a lifting of this basis. Thus $M=N+pM$. By Nakayama lemma we have $(1+p)M/N=0$. As $1+p$ is invertible in $\mathbb{Z}_p$ it is invertible in $B$. Thus $M=N$. In other words $M=(b_1,\dots,b_n)$. We will prove that the $b_i$'s form a basis. Suppose that there exist $a_1,\dots,a_n$ such that $\sum_{i=1}^{n}a_ib_i=0$. Thus working on $M/pM$ we deduce that $p\mid a_i$ thus for $1\leq i\leq 0$ we have $a_i=pa'_i$ for some $a_i'\in B$. Similarily we show that $a'_i=p a''_i$ for some $a''_i\in B$. We deduce that $a_i\in \bigcap_{n>0}p^nB$. By Artin-Rees lemma $\bigcap_{n>0}p^nB$ is the set of $x\in B$ such that $(1+p)x=0$. The only possibility is that $x=0$. Which concludes the lemma.