It’s a theorem of Brandl, Bubboloni, and Hupp that polynomials over Q which have roots modulo all primes are either linear or the product of at least two irreducible factors.
Is this still true if you replace “all primes” by “all but finitely many primes”?
The result in the original post was surely known to Kronecker. Indeed, let $f$ be a polynomial over $\mathbb{Z}$, and let $p$ run through the primes. Kronecker (1880) proved that the number of irreducible factors of $f$ equals the average number of zeros of $f\bmod p$, while $f\bmod p$ splits into linear factors for a positive density of primes $p$. See Lenstra-Stevenhagen: Chebotarev and his density theorem.
In particular, the answer to the OP's question is "yes", and one can even derive the following stronger statement. If $f\bmod p$ has a root for a set of primes $p$ of density exceeding $$1-\frac{\deg(f)-1}{\deg(f)!},$$ then $f$ has at least two irreducible factors.
