The proposed inequality $\|\log(RS)\|\le\|\log(R)+\log)S)\|$ does *not* hold for all $R,S\in\mathrm{SO}(3)$ for which $\log(RS)$, $\log(R)$, and $\log(S)$ are defined.  (In what follows, I'm taking the Frobenius norm to be $\|B\|^2 = \tfrac12\,\mathrm{tr}(B^TB)$, but this normalization clearly does not affect the argument.)

First, of all, notice that the function $B\mapsto \|\log(B)\|$ is actually a continuous function on $\mathrm{SO}(3)$, because of the easily checked identity
$$
\cos\bigl(\|\log(B)\|\bigr) = \tfrac12\bigl(\mathrm{tr}(B)-1\bigr)
$$
and the fact that $\cos^{-1}:[-1,1]\to[0,\pi]$ is a continuous, strictly decreasing function on $[-1,1]$.  If the desired inequality were actually true, then we would have the inequality
$$
\mathrm{tr}(RS) 
\ge 1+ 2\,\cos\bigl(\|\log(R)+\log(S)\|\bigr).
$$
Now consider the two matrices
$$
R = \exp\begin{pmatrix}0&-x&0\\x&0&0\\0&0&0\end{pmatrix}
\quad\text{and}\quad
S = \exp\begin{pmatrix}0&0&-y\\0&0&0\\y&0&0\end{pmatrix},
$$
where $0\le x,y < \pi$ (so that $R$, $S$, and $RS$ remain in the domain of the $\log$ function).  Then, if the above inequality held, it would imply that
$$
\cos x\cos y+\cos x + \cos y \ge 1 + 2\,\cos\left(\sqrt{x^2+y^2}\right).
$$
However, this is not true.  If you take $x=y$ in the above expression,
then, indeed, the inequality holds for small $x\ge0$, but there is a value
$x=y\approx 2.405181<\pi$ where both sides of this last inequality become equal and then the inequality reverses in the range $2.405182\le x=y <\pi$. 

Thus, the original proposed inequality does not hold.