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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 of the reduction (mod $p$) of $L$?" 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 $L^{G}$.

When $G = \langle u \rangle$ 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 $u$ 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 $p^n \langle \chi, 1 \rangle = p^{n-1}\langle {\rm Res}^{G}_{\langle u^{p} \rangle}(\chi),1 \rangle + (p^n - p^{n-1})\chi(u),$ 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 of the reduction (mod $p$) of $L$?" 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 $L^{G}$.

However, I will give an "algorithmic" description of how to proceed which might be helpful in the context of the original question (at least for cyclic $p$-groups, $p$ odd). As was discussed in the previous version of this question, it is possible to solve this problem in the case of a cyclic group of order $p$, even at the integral level. The description of what to do is easy. If $G$ is a cyclic group of order $p$, there are three isomorphism types of indecomposable $\mathbb{Z}_{p}G$-lattices. If $L$ is a $\mathbb{Z}_pG$-lattice, then $L$ has $a$ trivial indecomposable summands, $b$ indecomposable summands of rank $p-1$, and $c$ (projective) indecompsable summands of rank $p$. Here, $a$, $b$ and $c$ are respectively the number of Jordan blocks of size $1$,$p-1$ and $p$ in the reduction (mod p).$

Now suppose that $G = \langle u \rangle$ is cyclic of order $p^n$ ($p$ odd) and we have a $\mathbb{Z}_p G$-lattice $L$. Let $v$ denote a generator of the unique subgroup of order $p$ of $G$. We can determine the rank of the fixed sublattice $L^{\langle v \rangle}$, as just discussed. This is a pure submodule of the original $\mathbb{Z}_pG$-module. The minimum polynomial of $u$ on the quotient lattice $L/L^{\langle v \rangle}$ is $\Phi_{n}(x) = \frac{x^{p^n}-1}{x^{p^{n-1}}-1},$ because of the choice of $v$.

Let $N = \{ w \in L: w \Phi_n(x) = 0 \},$ a pure submodule of $L.$ Then the rank of $N$ is rank($L$) - rank($L^{\langle v \rangle}),$ which is determined by the reduction (mod $p$) of $L.$ Now $N + L^{\langle v \rangle}$ is a full (ie maximal rank) submodule of $L,$ and the quotient of $L$ by this submodule is a torsion module. But we can still form the quotient module $L/N$. Now $v$ acts trivially on $L/N$ by construction, so $L/N$ is "really" a module for the smaller group $G/\langle v \rangle$, and the rank of the $G$-fixed points on this module is the same as the rank of the $G$-fixed points of $L$. It might appear, then, that we are finished by an inductive argument in our quest to find the rank of the $G$-fixed points on $L.$ The issue, though, is in what sense we can claim to know the reduction (mod $p$) of $L/N$, and in what sense the submodule $N$ itself (rather than just its rank) is determined by the reduction (mod $p$) of $L.$ This is why I do not consider that this is a proper solution to the question.

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 $L^{G}$.

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 of the reduction (mod $p$) of $L$?" 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 $L^{G}$.

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 $M = L \otimes_{\mathbb{Z}_{p}} \mathbb{Q}_p$. 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 $p^n$, 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 $p^n \langle \chi, 1 \rangle = p^{n-1}\langle {\rm Res}^{G}_{\langle x^{p^{n-1}} \rangle}(\chi),1 \rangle + (p^n - p^{n-1})\chi(x),$ 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 determined the rank of $L^{G}$, the character is detemined completely, while if we know the character, we certainly know the rank of $L^{G}$.

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 $L^{G}$.