Let $R\subset S$ be commutative rings, $I<R$ an ideal and $M$ be an $R$-module. Suppose that

1) $R$ is Noetherian and $I$-adically complete.

2) $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)

3) $S$ is a flat $R$-algebra.

4) $S$ is $I$-adically complete

5) $M/IM$ is free module over $R/I$,

>Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?

I am only able to prove the above under the assumption that

6) $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$
in the following way:

Consider a resolution of $M$ in $R$-$\operatorname{Mod}$
$$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$
by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence 
$$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$
using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences
$$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$
by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude 
$\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel
of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$. 

Can one show this without assumption 6), but somehow to involve the second half of condition 1) and perhaps condtion 5)