= Solution
Complete multiplicativity and $f(p)=\pm1$ give, for $\operatorname{Re}s>1$,
$$
\boxed{F(s)=\prod_p(1-f(p)p^{-s})^{-1}.}
$$
The product is nonzero there; its absolutely convergent <logarithm> has terms $f(p)^mp^{-ms}/m$. Thus
$$
3\log\zeta(\sigma)+4\log|F(\sigma+it)|+\log|\zeta(\sigma+2it)|
=\sum_{p,m\ge1}\frac{3+4f(p)^m\cos(mt\log p)+\cos(2mt\log p)}{mp^{m\sigma}}.
$$
Each numerator is one of $3\pm4\cos\theta+\cos2\theta$, so it is nonnegative. Exponentiation gives the required lower bound one.
If $F(1+it)=0$ with $t\ne0$, its assumed holomorphy makes $|F(\sigma+it)|^4=O((\sigma-1)^4)$, or smaller for a higher-order zero. The zeta factor at real $\sigma$ grows at most as $(\sigma-1)^{-3}$, while the zeta factor near $1+2it$ stays bounded. The product tends to zero, a contradiction. Hence
$$
\boxed{F(1+it)\ne0\qquad(t\ne0).}
$$
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