Let $f \in \mathbb{Z}[x]$. Does $\bar{f}$ have as many roots in $\mathbb{F}_p$ as $f$ has in $\mathbb{C}$ for infinitely many primes $p$?

Yes. I will focus on the case that $f$ is irreducible, thus also separable, over $\mathbf Q$. (The general case can be deduced from this, the main case of interest.) For large primes $p$ the reduced polynomial mod $p$ is separable. There is a theorem in algebraic number theory that "in each number field, infinitely many primes split completely". This follows from the zeta-function of each number field having a simple pole at $s=1$. In the application of this to the field $K = \mathbf Q(a)$, where $f(a) = 0$, a large prime $p$ splitting completely in $K$ will be a prime $p$ for which $f \bmod p$ splits into distinct linear factors.

Yes. Chebotareff (really Frobenius here) density theorem says that each cycle type in the Galois group of $f$ occurs as the splitting type of $f$ modulo $p$ infinitely often. Your case corresponds to the cycle type of the identity element. The Chebotareff theorem says that this happens asymptotically once in $|Gal(f)|$ primes (the statement about infinite number of primes is actually much easier, and follows from the Chinese remainder theorem, but what the heck).