In the case of a cyclic $p$-group $G$, $p$ odd, it still seems unlikely to me that the reduction (mod $p$) of a $\mathbb{Z}_pG$-lattice $L$ will determine the isomorphism type of $M = L \otimes_{\mathbb{Z}_{p}} \mathbb{Q}_p$ in general. I have not found a counterexample so far, though they may well exist in the literature. However, I will record a couple of comments in case they are of use to someone else, (possibly in a positive direction if my intuition is wrong).
When $G = \langle x \rangle$ is cyclic of order $p^n$, ($p$ still odd), there are clearly just $n+1$ isomorphism types of irreducible $\mathbb{Q}_pG$-modules. These are the trivial module, and the representations obtained by representing $x$ respectively as the companion matrix of the irreducible polynomial $\frac{x^{p^m}-1}{x^{p^{m-1}}-1}$ for $1 \leq m \leq n.$
Let's label these as $V_0,V_1, \ldots,V_n$, where $V_0$ denotes the trivial module. Since $\mathbb{Q}_p$ has characteristic zero, the isomorphism type of $M$ is determined by the character $\chi$ it affords. As was known to E. Artin and R. Brauer, and is easily checked, in this situation, knowledge of the character afforded by $m$ is equivalent to knowing the dimension of the fixed-point space of $x^{p^j}$ on $M$ for each $j$ with $ 0 \leq j \leq n.$ Clearly $\chi$ determines the dimension of these fixed point spaces. On the other hand, these dimensions determine $\chi$ inductively because of the fact that $p^n \langle \chi, 1 \rangle = p^{n-1}\langle {\rm Res}^{G}_{\langle x^{p} \rangle}(\chi),1 \rangle + (p^n - p^{n-1})\chi(x),$ since $\chi$ is rational valued.
Hence (for $p$ odd and a cyclic $p$-group $G$), the question is equivalent to " can we determine the rank of the fixed sublattice $L^{G}$ solely from knowledge its reduction (mod $p$)?" For if we were working inductively, we could assume that we knew how the character restricted to every proper subgroup of $G$, since we certainly know the reduction $(mod $p$)$ of all these restrictions. Hence, as above, if we can determine the rank of $L^{G}$, the character is detemined completely, while if we know the character, we certainly know the rank of $L^{G}$.