Suppose that we do not return balls to the box. The plain probability that the $i$-th ball selected is white is $\frac{k}{N}$, exactly like in the case of returning the ball to the box.
One way of seeing this is to number the balls, white from $1$ to $k$, black from $k+1$ to $N$. Imagine that we draw the balls, one at a time, until all the balls are gone. All permutations of the labels are equally likely, and the fraction of these permutations for which the $i$-th ball drawn is white is $\dfrac{k}{N}$. (It can take a while until this fact becomes "obvious"!)
But in drawing one at a time, there are other probabilities that we can compute, for example the probability that the $3$rd ball drawn is white given that the first two were white. This conditional probability is not $\dfrac{k}{N}$, it is $\dfrac{k-2}{N-2}$ (except in trivial cases, like $k=1$).
Suppose again that we draw the balls, one at a time, until they are all gone. The conditional probability that the $3$rd ball drawn is white, given that the last two balls drawn (of the $N$) are white is also $\dfrac{k-2}{N-2}$. So it is not the temporal order of the drawing that matters in evaluating the conditional probability.
When we calculate a conditional probability, it is not the act of receiving information that matters, it is the act of using the information to calculate a conditional probability, that is, to restrict the sample space.
When one does replacement, conditional and unconditional probabilities are the same, since the sample space is unchanged.
Best Answer
The number of ways to choose two black and one white square is ${32 \choose 2}32$ with the first factor from choosing the two black squares and the second from choosing the white square. Two white and one black is the same by symmetry. There are ${64 \choose 3}$ ways to choose three squares, so the chance you want is $$\frac {2\cdot {32 \choose 2}\cdot 32}{{64 \choose 3}}=\frac {16}{21}$$ which is nicely less than $1$.