We can actually solve this recurrence exactly and not just for powers of three, at any rate one that is very close. Consider
$$T(n) = 2 \; T(\lfloor n/3\rfloor) + \lfloor\log_3 n\rfloor+1$$
with $T(0) = 0$ and which has $T(1) = 1.$ The constant one has lower order than the logarithm so the complexity is the same as in the original query. (Note $|\lfloor\log_3 n\rfloor-\log_3 n | < 1$ and that $|\lfloor\log_3 n\rfloor+1-\log_3 n | < 1$ so adding in one introduces the same error as using the floor alone.)
The key observation is that the answer does not depend on $n$ but only on the number of digits in base three that it has, i.e. the logarithm.
If $$n= \sum_{k=0}^{\lfloor\log_3 n\rfloor} d_k 3^k
\quad\text{then}\quad
T(n) = \sum_{k=0}^{\lfloor\log_3 n\rfloor} 2^k \; (\lfloor\log_3 n\rfloor+1-k)
= 4 \times 2^{\lfloor\log_3 n\rfloor} -\lfloor\log_3 n\rfloor -3.$$
This formula is exact for all $n.$
Now clearly the dominant term is the exponential one which gives the following complexity:
$$\Theta\left(2^{\lfloor\log_3 n\rfloor}\right) =
\Theta\left(3^{\log_3 2 \times \lfloor\log_3 n\rfloor}\right) =
\Theta(n^{\log_3 2}).$$
This is of course the same as what the Master theorem would produce. Here is another calculation of the same type that is a bit more advanced.
