No, we cannot.
Let m and h be the positions of the minute and second hands according to where the minute markings are, so h,m are in [0,60). At all times, h mod 5 = m/12. If we can't tell the time, that means that additionally, m mod 5 = h/12; and finally, we must have m not equal to h, because if they're in the same place, it doesn't matter which is which. The first two equations will be satisfied if and only if 12((12(h mod 5))mod 5)-h = 0. You can run a computer program to solve this, but you can also use the Intermediate Value Theorem: the expression on the left, viewed as a function, is continuous except at multiples of 5/12, so we can make it continuous by restricting it to [5/12, 9/12]. You can easily calculate that the expression is negative at h = 5/12 and positive at h = 9/12, so by the IVT, it must be equal to zero somewhere in between. We know that the hands are not overlapping there, since that only happens once per hour and has already occurred this hour at 12:00. Thus, there is some time between 12:25 and 12:45 when we can't tell what time it is.
This isn't necessarily the only time when this is the case—indeed, there must be at least one more, obtained by switching the positions of the minute and hour hands (after all, that's why it's indistinguishable). We can find more by finding other solutions to the equation. However, I'm fairly certain that there are only finitely many solutions, so most of the time, we can in fact tell what time it is.