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.

[**added later**] For $f_a(x) := \prod_{k=1}^\infty \cos(x/a^k)$
with $a > 2$, the same construction does work if $a$ is an integer
(consider $x = a^n \pi$), and more generally if $a$ is a
Pisot-Vijayaraghavan number (which need not exceed $2$,
e.g. if $a = (1+\sqrt5)/2$ then there are arbitrarily large $x$
such that $f_a(x)$ remains bounded away from zero).