Yes. For example, we can take $f(x) = x^{2007} - x^{2006} + 1$.
It suffices to show that $(m, f^k(m)) = 1$ for any positive integer $k$, since we can replace $m$ by any $f^j(n)$. Consider any prime $p$ dividing $m$. Then we have $f(m) \equiv 0 \pmod{p} \equiv -0 + 1 = 1 \pmod{p}$. Whenever $f^i(m) \equiv 1 \pmod{p}$, we have
$$f^{i+1}(m) \equiv 1 - 1 + 1 \equiv 1 \pmod{p}.$$
Thus, $f^k(m) \equiv 1 \pmod{p}$ for any $k \ge 1$ by induction. This shows $p \nmid f^k(m)$, and hence $(m, f^k(m)) = 1$. This completes the proof.