Do the base 3 digits of $2^n$ avoid the digit 2 infinitely often -- what is the status of this problem?

As of a few months ago, the status of the problem was: still unsolved. See the slides Jeff Lagarias put up from a talk he gave in September 2009: http://www.math.lsa.umich.edu/~lagarias/TALK-SLIDES/ternary-fields-2009sep.pdf

An older reference is http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.65.6976&rep=rep1&type=pdf (Brian Hayes, Third Base, American Scientist) which says the problem was still open in late 2001; also that Ilan Vardi searched up to $2^{6973568802}$ without finding any 2-less powers of 2 (other than $2^2$ and $2^8$).

... and $\sum (2/3)^{n\log 2/\log 3} < \infty$ so (if things were random) we would expect only finitely many such occurrences by Borel-Cantelli easy direction.

But of course proving this is anything like random is far too hard for today's tools, I think.

Something way easier that one can prove is that your sequence is disjunctive, which is much weaker than normality. This is true since for every string of digits $k$ you can find a power of two with a base 3 expansion that starts with k.

I don't know much about normality except that there are many conjectures (for example that every irrational algebraic number is normal) and no techniques to answer such questions except in trivial cases. I believe the current state is similar for simply normal numbers.

Another question that is similar in spirit and that was answered recently is Stolarsky's conjecture that says $$\liminf_{n\to \infty}\frac{s_q(n^k)}{s_q(n)}=0$$ where $s_q(\cdot)$ is the sum of digits in base $q$. Intuitively it is hard to come up with examples that $s _3(2^{nk)} < s_3(2^k)$ for most $k$, let alone that the $\liminf$ is zero. However this question is much weaker than the one you ask since the sum of digits puts very few restraints on the digits themselves.