After looking at Geoff's answer and §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of $C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane).

So, here is the counterexample: Consider the following two matrices in $\mathbb Q_3^{10\times 10}$: $$ A=\left(\begin{array}{rrrrrrrrrr} 1&0&0&0&1&0&0&0&0&0\newline 0&0&0&1&1&0&0&0&0&0\newline 0&1&0&0&1&0&0&0&0&0\newline 0&0&1&0&1&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ and $$ B = \left(\begin{array}{rrrrrrrrrr}% 1&0&0&0&0&0&0&0&0&0\newline 0&0&0&1&0&0&0&0&0&0\newline 0&1&0&0&0&0&0&0&0&0\newline 0&0&1&0&0&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ Then $A$ and $B$ are conjugate over $\mathbb Q_3$ but their reductions to $\mathbb F_3$ are not conjugate. Moreover, both have finite order (their order is $9$).

To see that their reductions mod 3 are not conjugate just compute the rank of $A-\textrm{id}$ and $B-\textrm{id}$ in $\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and $7$ in the second, so they cannot be conjugate.

To see that they are conjugate over $\mathbb Q_3$ just compute minimal and characteristic polynomial. The minimal polynomial is $x^9-1$ in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomial is $(x-1)(x^9-1)$ in both cases.

Edit: Sorry, just realized this is in in the wrong direction. What was asked for was an example of matrices over $\mathbb Z_p$ which are conjugate over $\mathbb F_p$ but not over $\mathbb Q_p$; I swapped $\mathbb Q_p$ and $\mathbb F_p$, but then of course the assertion is clear anyway (I'll look at it again later).

After looking at Geoff's answer and §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of $C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane).

So, here is the counterexample: Consider the following two matrices in $\mathbb Q_3^{10\times 10}$: $$ A=\left(\begin{array}{rrrrrrrrrr} 1&0&0&0&1&0&0&0&0&0\newline 0&0&0&1&1&0&0&0&0&0\newline 0&1&0&0&1&0&0&0&0&0\newline 0&0&1&0&1&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ and $$ B = \left(\begin{array}{rrrrrrrrrr}% 1&0&0&0&0&0&0&0&0&0\newline 0&0&0&1&0&0&0&0&0&0\newline 0&1&0&0&0&0&0&0&0&0\newline 0&0&1&0&0&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ Then $A$ and $B$ are conjugate over $\mathbb Q_3$ but their reductions to $\mathbb F_3$ are not conjugate. Moreover, both have finite order (their order is $9$).

To see that their reductions mod 3 are not conjugate just compute the rank of $A-\textrm{id}$ and $B-\textrm{id}$ in $\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and $7$ in the second, so they cannot be conjugate.

To see that they are conjugate over $\mathbb Q_3$ just compute minimal and characteristic polynomial. The minimal polynomial is $x^9-1$ in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomial is $(x-1)(x^9-1)$ in both cases.

After looking at Geoff's answer and chapter §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of $C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane).

So, here is the counterexample: Consider the following two matrices in $\mathbb Q_3^{10\times 10}$: $$ A=\left(\begin{array}{rrrrrrrrrr} 1&0&0&0&1&0&0&0&0&0\newline 0&0&0&1&1&0&0&0&0&0\newline 0&1&0&0&1&0&0&0&0&0\newline 0&0&1&0&1&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ and $$ B = \left(\begin{array}{rrrrrrrrrr}% 1&0&0&0&0&0&0&0&0&0\newline 0&0&0&1&0&0&0&0&0&0\newline 0&1&0&0&0&0&0&0&0&0\newline 0&0&1&0&0&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ Then $A$ and $B$ are conjugate over $\mathbb Q_3$ but their reductions to $\mathbb F_3$ are not conjugate. Moreover, both have finite order (their order is $9$).

To see that their reductions mod 3 are not conjugate just compute the rank of $A-\textrm{id}$ and $B-\textrm{id}$ in $\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and $7$ in the second, so they cannot be conjugate.

To see that they are conjugate over $\mathbb Q_3$ just compute minimal and characteristic polynomial. The minimal polynomial is $x^9-1$ in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomials are $(x-1)(x^9-1)$ in both cases.

After looking at Geoff's answer and §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of $C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane).

So, here is the counterexample: Consider the following two matrices in $\mathbb Q_3^{10\times 10}$: $$ A=\left(\begin{array}{rrrrrrrrrr} 1&0&0&0&1&0&0&0&0&0\newline 0&0&0&1&1&0&0&0&0&0\newline 0&1&0&0&1&0&0&0&0&0\newline 0&0&1&0&1&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ and $$ B = \left(\begin{array}{rrrrrrrrrr}% 1&0&0&0&0&0&0&0&0&0\newline 0&0&0&1&0&0&0&0&0&0\newline 0&1&0&0&0&0&0&0&0&0\newline 0&0&1&0&0&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right) $$ Then $A$ and $B$ are conjugate over $\mathbb Q_3$ but their reductions to $\mathbb F_3$ are not conjugate. Moreover, both have finite order (their order is $9$).

To see that their reductions mod 3 are not conjugate just compute the rank of $A-\textrm{id}$ and $B-\textrm{id}$ in $\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and $7$ in the second, so they cannot be conjugate.

To see that they are conjugate over $\mathbb Q_3$ just compute minimal and characteristic polynomial. The minimal polynomial is $x^9-1$ in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomial is $(x-1)(x^9-1)$ in both cases.