No. Let $\omega(p)$ be the number of roots of $f$ modulo $p$. Clearly, for any finite set $S$, the upper asymptotic density of your set is bounded by $\prod_{p \in S} (1-\omega(p)/p)$. (Because the probability that $p \nmid f(n)$ is $1-\omega(p)/p$, these probabilities are independent for distinct primes, and $f(n)$ only equals $p$ finitely many times.) We have $ \prod_{p \in S} (1-\omega(p)/p) \leq \exp (- \sum_{p \in S} \omega(p)/p)$.
But the Cebotarov (or Frobenius) density theorem gives that $\sum \omega(p)/p$ diverges to $\infty$, so we may take a finite set $S$ large enough to make $\sum_{p \in S} \omega(p)/p$ greater than any specified $N$.
I'll mention how this fits into the Bateman-Horn conjecture. That says that the density should go to $0$ like $\prod \frac{p-\omega(p)}{p-1} \cdot \frac{x}{\log f(x)}$, where the cool thing is that the product converges to a nonzero number if and only (1) $f$ is irreducible and (2) $\omega(p) \neq p$ for any $p$. But all I need to answer your question is an upper bound of $\prod_{p \leq N} \frac{p-\omega(p)}{p} \cdot x$.