Answer: Yes, for example $Q(x) = (2x^3 + 1)(x^2 - x + 1)$.
Clearly $Q(x)$ has no rational roots. We will show that the only prime excluded by $Q(x)$ is $p=2$.

Observation 1: For any prime $p$ satisfying $p \equiv 1 \pmod 3$ there exists an integer $n$ such that $n^2 - n + 1 \equiv 0 \pmod p$.
Proof. Let $\omega$ be a primitive root modulo $p$ and $n = -\omega^{(p-1)/3}$. Then $n^3 \equiv -1 \pmod p$ and $n \not\equiv -1 \pmod p$. Now since $n^3 + 1 = (n+1)(n^2 - n + 1)$ we get $n^2 - n + 1 \equiv 0 \pmod p$.

Observation 2: For any odd prime $p$ satisfying $p \equiv 2 \pmod 3$ there exists an integer $n$ such that $2n^3 + 1 \equiv 0 \pmod p$.
Proof. Let $a$ be an integer satisfying $2a+1 \equiv 0 \pmod p$. Since $3 \mid p-2$ we get $3 \mid 2p-1$. Then for the integer $n = a^{(2p-1)/3}$ by Fermat's little theorem we have $n^3 \equiv a^{2p-1} \equiv a \pmod p$.

Therefore, $2n^3 + 1 \equiv 2a + 1 \equiv 0 \pmod{p}$.
$Q(x)$ does not exclude $p=3$ since $3 \mid Q(2)$. By Observation 1 any prime $p \equiv 1 \pmod 3$ is not excluded by $Q(x)$. By Observation 2 any odd prime $p \equiv 2 \pmod 3$ is not excluded by $Q(x)$. Since both $2n^3 + 1$ and $n^2 - n + 1$ are always odd numbers $p=2$ is excluded by $Q(x)$. We are done.