This is only an observation, but perhaps it would be helpful. The problem is equivalent to the case $M$ being a cyclic prime module, i.e $M=S/P $ for some prime ideal $P$.

That is because if we take a prime filtration $\mathcal F$ whose quotient has only associated primes of $M$, then any regular element has to be outside all of those primes. By induction on the length of $\mathcal F$, $M/xM$ has a filtration with quotients of the forms $S/(P_i,x)$, where $P_i \in Ass(M)$. So if $M$ is not almost clean, then one of the quotients can't be almost clean.

So now the question really is: Is there a domain $R$, and an element $x\in R$ such that $R/xR$ is not almost clean? I think there are, but proving they work might involve some K-theoretic arguments (see this question by Steven Landsburg).

This is only an observation, but perhaps it would be helpful. The problem is equivalent to the case $M$ being a cyclic prime module, i.e $M=S/P $ for some prime ideal $P$.

That is because if we take a prime filtration $\mathcal F$ whose quotient has only associated primes of $M$, then any regular element has to be outside all of those primes. By induction on the length of $\mathcal F$, $M/xM$ has a filtration with quotients of the forms $S/(P_i,x)$, where $P_i \in Ass(M)$. So if $M$ is not almost clean, then one of the quotients can't be almost clean.

So now the question really is: Is there a domain $R$, and an element $x\in R$ such that $R/xR$ is not almost clean? I think there are, but proving they work might involve some K-theoretic arguments (see this question by Steven Landsburg).