This is an elaboration on Tyler Lawson's comment: For $A \in \mathrm{Mat}_{n \times n}(\mathbb{Q})$, the following are equivalent:

(0) There is a nonzero integer $M$ so that $M \cdot A^r$ has integer entries for all $r \geq 0$.

(1) $A = g B g^{-1}$ for some $B \in \mathrm{Mat}_{n \times n}(\mathbb{Z})$ and $g \in GL_n(\mathbb{Q})$

(2) All the coefficients of the characteristic polynomial $\det(t \mathrm{Id} - A)$ are integers.

Proofs: $(0) \implies (1)$. Let $\Lambda$ be the subgroup of $\mathbb{Q}^n$ generated by $A^r \mathbb{Z}^n$ for all $r \geq 0$. Then $\frac{1}{M} \mathbb{Z}^n \subseteq \Lambda \subseteq \mathbb{Z}^n$ so $\Lambda$ is a free abelian group of rank $n$. Choosing a basis for $\Lambda$ places $A$ in the required form.

$(1) \implies (2)$ Obvious.

$(2) \implies (0)$ Let the characteristic polynomial be $t^n + b_{n-1} t^{n-1} + \cdots b_1 t + b_0$. Then, by the Cayley-Hamilton theorem, 
$$A^{r+n} = - \left( b_{n-1} A^{r+n-1} + \cdots + b_1 A^{r+1} + b_0 A^r \right).$$
So, for $s \geq n$, the matrix $A^s$ is in the integer span of $A^0$, $A^1$, ..., $A^{n-1}$. Choosing $M$ large enough to clear the denominators of these finitely many matrices proves the result.

Note that description (1) makes it easy to find many examples.