Let $A$ be a commutative ring (not necessarily Noetherian), $I=(f_1, f_2, \dots, f_n) \subseteq A$ a finitely generated ideal that is generated by a regular sequence. Let $M$ be an $A$-module, and assume that both $M$ and $A$ are $I$-adically complete. Finally, suppose further that $M$ is $I$-completely flat (sometimes also called $I$-adically flat), meaning that $\mathrm{Tor}_{>0}^{A}(A/I, M)=0$ and $M/IM$ is a flat $A/I$-module.

My question is

If $x \in A$ is a non-zero divisor of $A$, is $x$ necessarily a non-zero divisor on $M$?

This seems true at least under some further assumptions on $x$, such as $xA \cap I^k=xI^k$ for all $k$, but I am actually interested in the somewhat perpendicular case when $x \in I$, or better yet, the general question whether such $I$-completely flat modules are torsion-free.

Some details:

One can look at the exact sequences $$0 \rightarrow K_n/I^n \rightarrow A/I^n \stackrel{x}{\rightarrow} A/I^n $$ where $K_n=(I^n: x)$. Taking $\varprojlim_n,$ from the fact that $x$ is a non-zero divisor on $A$ it follows that $\varprojlim_n K_n/I^n=0$. It can be shown that the $I$-complete flatness is equivalent to $I^n$-complete flatness. Thus, applying $-\otimes_AM=-\otimes_{A/I^n}M/I^nM$ to the above exact sequence, the sequence $$0 \rightarrow K_nM/I^nM \rightarrow M/I^nM \stackrel{x}{\rightarrow} M/I^nM$$ remains exact, and again, taking $\varprojlim_n$ one obtains that $Ann_M(x)=\varprojlim_n K_nM/I^nM$. So the question is, is $\varprojlim_nK_n/I^n=0$ anough to guarantee $\varprojlim_nK_nM/I^nM =0$ in this situation?