Hint
Bayes' theorem. It will be helpful to consider the $4$ kinds of cubes: from the corner of the initial $8\times8$ cube, or from an edge, or from the interior of a face, or from inside the cube. Count how many there are of each kind (and how many painted faces they have), then the probability to roll a blank bottom for each case. Then apply Bayes' theorem.
Detailed solution
(note that the number of unpainted cubes is not $6\times64=384$ but $6^3=216$)
Corner cubes: $8$ such cubes, with $3$ faces painted (hence probability $3/6=1/2$ to roll a blank bottom.
Edge cubes (and not corner): $6$ on each edge, and there are $12$ edges, hence $72$ such cubes, with $2$ painted faces each. Probability to roll a blank bottom: $4/6=2/3$.
Face cubes (and not corner nor edge): $6\times6=36$ on each face, and there are $6$ faces, hence $216$ such cubes, with $1$ painted face each. Probability to roll a blank bottom: $5/6$.
Inner cubes: $6^3=216$ such cubes, and they have no painted face. Probability to roll a blank bottom: $1$.
Quick check: $216+216+72+8=512=8^3$.
What is the probability to roll a blank bottom (called event "blank" below)?
$$P(blank)=P(corner)\cdot P(blank|corner)+P(edge)\cdot P(blank|edge)\\+P(face)\cdot P(blank|face)+P(inner)\cdot P(blank|inner)\\=\frac{8}{512}\cdot\frac{3}{6}+\frac{72}{512}\cdot\frac{4}{6}+\frac{216}{512}\cdot\frac{5}{6}+\frac{216}{512}\cdot\frac{6}{6}=\frac78$$
Probability to have an unpainted cube given that the bottom is blank:
$$P(unpainted|blank)=\frac{P(unpainted\ and\ blank)}{P(blank)}=\frac{P(unpainted)}{P(blank)}=\frac{216/512}{7/8}=\frac{27}{56}\approx 0.48$$
If it’s one of the cubes in the centres of the faces, there is only one chance in six that the red face is on the bottom, but if it’s the central cube, any of the six faces could be on the bottom. To put it a bit differently, the $7$ cubes have a total of $42$ faces, all of which are equally likely to be on the bottom. If you don’t see a red face, you know that the face on the bottom is either one of the $6$ faces of the central cube or the one red face of one of the other $6$ cubes. Thus, it’s one of $12$ faces, and your sample space actually has size $12$. In $6$ of those $12$ cases the bottom face is red, so the probability is actually $\frac12$.
Best Answer
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $3\times 3\times 3$ cube, you paint a total of $$ 9 \cdot 6 = 54 $$ faces (each face of the large cube consists of a $3\times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of $$ 3 \cdot 8 = 24 $$ painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e. $$ P(\text{the painted face is from a corner}) = \frac{\text{number of successful outcomes}}{\text{total number of outcomes}} = \frac{24}{54} = \frac{4}{9}. $$