Suppose some uniformly continuous (under the 2-adic topology) function $f:\Bbb Z[\frac16]\to\Bbb Z[\frac16]$ converges under composition with itself to $0$ (in the 2-adic topology) for all inputs, and also satisfies $f(0)=0$.
Also let $f$ preserve the $3$-adic valuation, i.e. $\lvert f(x)\rvert_3=\lvert x\rvert_3$
Is there some reason why such a function can approach $0$ only through some sequence terminating in $2^m3^n:m\to\infty$?
I can't show that some counterexample is a counterexample, but then my skills are sadly lacking.
I'm not sure that I understood all the conditions, but I think that the following function may be of interest here: $$ f(x)=10x. $$ Denoting the $n$-fold iterate by $f^{[n]}$. Because $2\mid 10$ we have $$|f^{[n]}(x)|_2=(1/2)^n|x|_2$$ for all $x$. Therefore, for all $x$, $$\lim_{n\to\infty}f^{[n]}(x)=0$$ with respect to the $2$-adic metric. For the same reason $f$ is uniformly continuous w.r.t. the $2$-adic metric.
I'm uncertain what you mean by can only approach zero through some sequence terminating in $2^m3^n$ means, but starting with the initial point $x=1$ the sequence $(f^{[n]}(x))_{n\in\Bbb{N}}$ has no entries of the form $2^n3^m$.
Because $\gcd(3,10)=1$ we have $|f(x)|_3=|x|_3$ for all $x$.