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.