Let $A \in M_n(\mathbb{Z}_p)$ be a nonsingular matrix which is nilpotent mod $p$, so $A^r \in pM_n(\mathbb{Z}_p)$ for some $r$. Then $\mathbb{Z}_p[[T]]$ acts on $\mathbb{Z}_p^n$ with $T$ acting by $A$. If I take $M/pM$, or equivalently tensor with $\mathbb{F}_p[[T]]$, then since $\mathbb{F}_p[[T]]$ is a local ring I get a decomposition of the form $\bigoplus_i \mathbb{F}_p[[T]]/(T^{n_i})$. However, the structure theorem for modules over the Iwasawa algebra tells us that there is a quasi-isomorphism (i.e. a $\mathbb{Z}_p[[T]]$-module map with finite kernel and cokernel) $$M \to \bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$$ where the $f_i$ are either distinguished polynomials (there is no free part and no $p$-torsion do to the initial assumptions on $M$). If I take the module $\bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$ and reduce mod $p$ to get an $\mathbb{F}_p[[T]]$-module, then it's already in the form $\bigoplus_i \mathbb{F}_p[[T]]/(\bar{f}_i(T)^{n_i})$ (where the $\bar{f}_i$'s are just powers of $T$ since the $f_i$ were distinguished polynomials).

My question is, is the $\mathbb{F}_p[[T]]$-module I get by reducing $\bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$ mod $p$ the same as the one I got by reducing $M$ mod $p$, and then finding its decomposition? So in some sense I'm asking whether the operations "decompose my module" and "reduce mod $p$" commute. Note that this decomposition holds more information than the $\lambda$-invariant of $M$.

As Chris Wuthrich points out in a comment responding to an earlier version of this question where I only assumed $M$ is a finitely-generated torsion module which is free when restricted to a $\mathbb{Z}_p$-module, the answer is "no" under these conditions. However, the setup above gives some additional conditions, and hopefully these are enough to make the above true.

Note: I'm aware that the quasi-isomorphism from $M$ to the decomposed module will in general NOT become an isomorphism after tensoring the two modules with $\mathbb{F}_p[[T]]$, but I'm asking whether they will nonetheless be isomorphic by a different map.