Here is a discussion only assuming that $R$ is Cohen-Macaulay. Let $G$ denote the Grothendick group of all finitely generated $R$-modules. 

Fix a regular element $a$ and consider the function $$f(M) = \dim_k Tor_0(M,R/aR)- \dim_k Tor_1(M,R/aR)= \dim_k M/aM - \dim_k (0:_Ma)$$

 For torsion-free module (or just Cohen-Macaulay) $M$, note that $f(M) = \dim_k M/aM$. Also, note that $f(N)=0$ if $N$ is torsion. So, for example, $f(m)=f(R)$ if $m$ is a maximal prime.

Thus $f$ induces a group homomorphism $\bar G \to \mathbb Z$, where $\bar G$ is $G/ker(f)$. And $\bar G$ is generated as an abelian group by the classes of $[R/P]$, $P$ is a minimal prime.  

So, we have $f(M) = \mu f(R)$ if $[M]=\mu [R]$ in $\bar G$. This happens always if $R$ is a domain an $\mu$ is rank. In general one needs to understand $\bar G$ and $f(P)$ when $P$ is a minimal prime.

To be more precise, if $R$ is reduced, then $[R] = \sum_1^s [R/P_i]$ and $[M]= \sum a_i[R/P_i]$ in $\bar G$, where $P_i$ are minimal primes of $R$ and $a_i$ is the rank of $M_{P_i}$. So if we let $\mu= \min\{a_i\}$ or $\mu = \max\{a_i\}$ we get inequalities in either direction. If it happens that $f(P_i)$ is a constant then $\mu = \sum a_i/s$ works for equality, generalizing the domain case.