No. If $x = 3^n \pi$ then $|f(x)| = f(\pi) \neq 0$ (numerically it's about $0.466$), and $3^n \pi$ can be arbitrarily large.
Such a construction fails for $\prod_{k=1}^\infty \cos(x/2^k)$ because if some $x/2^k$ is nearly $\pi$, or more generally some odd multiple of $\pi$, then $x/2^{k+1}$ is nearly a half-integral multiple of $\pi$ and thus has a very small cosine.