You might be asking about four separate types of modules: 

* irreducible Z[G] modules,
* Z-forms of irreducible Q[G] modules,
* indecomposable Z[G] modules, or
* indecomposable Z[G] modules that are finitely generated and free as Z-modules.

I'll assume the last is the main concern.

The irreducible modules of ZS<sub>3</sub> are all finite and have an elementary abelian p-group as their additive group.  For p=2,3 there are 2 each, and for p>3, there are 3 each.

The irreducible CS<sub>3</sub> modules are all realizable over Q.  Every such module may be realized over Z, but for general finite G and possibly reducible modules, a QG module may have multiple inequivalent Z-forms.  I think for irreducible modules over S3, they are all unique.

Indecomposable ZS<sub>3</sub> modules up to isomorphism are more complicated than the human mind can possibly comprehend.  Indeed, even those in which S_3 acts as the identity are much too complex.  Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).

The indecomposable ZS<sub>3</sub> modules that are free as Z-modules are classified in:

Lee, Myrna Pike.  "Integral representations of dihedral groups of order 2p."
Trans. Amer. Math. Soc. 110 (1964) 213–231.
<a href="http://www.ams.org/mathscinet-getitem?mr=156896">MR 156896</a>
<a href="http://dx.doi.org/10.2307/1993702">doi:10.2307/1993702</a>

There are 10 of them, and the Krull-Schmidt theorem fails for them.  Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique.  In other words, integral representations of even very small groups are quite complicated.