At least in Faltings's setting, $m$ is the maximal ideal of a non-Noetherian valuation domain $V$.  If we let $S$ be the non-zero elements of $m$, then the localization of $V$ at $S$ will
be the field of fractions $K$ of $V$.    The category of $K$-modules (i.e. $K$-vector spaces)
*is* obtained as a Serre quotient of $V$-modules, but one quotients out modules which are
killed by *some* non-zero element of $V$, while the category of amost modules is obtained 
by quotienting out by a much smaller category, namely the modules which are killed by *every* element of $m$.  

In other words, I think you have misinterpreted the meaning of $m$-torsion, at least in so far it is used in the context of Faltings's "almost commutative algebra".

Added in response to the comment below:  Let me not use the word $m$-torsion anymore, since it can be interpreted in different ways, and seems to be causing confusion.

Let $\mathcal M$ be the category of $V$-modules.  Here are two Serre subcategories of $\mathcal M$:

1. $\mathcal C$, the category of almost zero modules, i.e. modules for which $m M = 0$.
This is a Serre subcategory because $m^2 = m$ in the context of Faltings's theory.

2. $\mathcal C'$, the category of torsion modules, i.e. modules such that $x M = 0$
for *some* non-zero $x \in m$.  This is a Serre category just because $V$ is a domain.

Clearly $\mathcal C$ is contained in $\mathcal C'$, but they are far from equal!

The quotient $\mathcal M/\mathcal C$ is the category of almost modules.  The quotient
$\mathcal M/\mathcal C'$ is equivalent to the category of $K$-vector spaces
(the equivalence being given by applying the functor $K\otimes\text{--}$, where $K$
is the fraction field of $V$).

In the statement of the question, you (mistakenly) define almost modules to be $\mathcal M/\mathcal C'$, and (correctly) conclude that this is equivalent to $K$-vector spaces.  Once you replace this mistaken definition with the correct definition of almost modules (i.e. $\mathcal M/\mathcal C$), you will see that the resulting category is *not* equivalent to $K$-vector spaces.