The answer is no. Here is a counter-example
$$left( begin{matrix} 0 & -5 \\ 1 & 0 end{matrix} right) quad mbox{and} quad left( begin{matrix} -1 & -3 \\ 2 & 1 end{matrix} right).$$
Both of these matrices have characteristic polynomial $x^2+5$. For $p neq 2$, $5$, this polynomial has no repeated factors so any matrices with this polynomial are similar. By the same argument, they are similar over $mathbb{Q}$. By brute force computation, they are also similar at $2$ and $5$.
However, they are not similar over $mathbb{Z}$. Consider $mathbb{Z}^2$ as a module for $mathbb{Z}[t]$ where $t$ acts by one of the two matrices above. Both of these matrices square to $-5$, so these are in fact $mathbb{Z}[sqrt{-5}]$-modules. If the matrices were similar, the similarity would give an isomorphism of $mathbb{Z}[sqrt{-5}]$-modules. But these are not isomorphic: the former is free on one generator while the latter is isomorphic to the ideal $langle 2, 1+ sqrt{-5} rangle$.
In general, the way to classify similarity of matrices over $mathbb{Z}$ is the following: If the matrices do not have the same characteristic polynomial over $mathbb{Q}$, they are not similar. If they do, let $f$ be the characteristic polynomial and let $R=mathbb{Z}[t]/f(t)$. Then your matrices give $R$-modules, and the matrices are similar if and only if the $R$-modules are isomorphic. If $R$ is the ring of integers of a number field, then $R$-modules are classified by the ideal class group. In general, they are related to the ideal class group, but there are various correction factors related to how $R$ fails to be the ring of integers of its fraction field (or how it fails to be a domain at all). I don't know the details here.
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