$P(x)= \frac{x(x+1)}{2} +1$.
It is easy to see that $P^{n+1}(0) > P^n(0)$ and $P$ maps the integers into the integers.
But I think (didn't check it, migth be one of these facts which are obvious but wrong) that
$$P^(n)(x) = \frac{1}{2^{m} x^{2^n}+....\notin \ZZ $$
where $m$ is probably $m=2^n+1$.
The right question to ask migth be if $f$ maps the integers into the integers....
EDIT: P.S. The answer with the rationals turns out to be true, I think (my algebra is rusty):
Let $P$ be such a polynomial, and let $m$ be the degree of $P$. Then using teh Lagrange interpolation formula, you can reconstruct $P(x)$ from $m+1$ distinct integer values of the type $P^{(k)}(0)$, and since all of these are integers, all the coefficients are rational.