Connection between $\mathbb{Q}_p[G]$ and $\mathbb{Z}_p[G]$ In this post there was the comment, that having $\mathbb{Q}_p[G]$ modules, it is possible to construct $\mathbb{Z}_p[G]$ modules. How is it possible to find out when there is a bijection between simple $\mathbb{Q}_p[G]$ modules and indecomposable  $\mathbb{Z}_p[G]$ modules? (Maybe without calculation the characters)

There will almost never be a bijection between simple $\mathbb Q_p[G]$-modules and indecomposable $\mathbb Z_p[G]$-modules.
For one thing, there are only finitely many isomorphism classes of the former, whereas, unless $G$ has a cyclic $p$-Sylow subgroup, there are infinitely many isomorphism classes of the latter.
(One basic point is that $M$ being an indecomposable $\mathbb Z_p[G]$-module
does not imply that $M\otimes_{\mathbb Z_p} \mathbb Q_p$ is simple over $\mathbb Q_p[G]$.)
Even if we restrict attention to indecomposable $\mathbb Z_p[G]$-modules $M$
for which $V:= M\otimes_{\mathbb Z_p} \mathbb Q_p$ is simple (i.e. $G$-invariant $\mathbb Z_p$-lattices in irred. reps. over $\mathbb Q_p$), there won't be such a bijection in general.
As noted in my comment on the question linked in your post, $M$ is determined up to isomorphism by $V$ only in the special case when $M/pM$ is irred. as a $G$-rep'n.  
