All charges don't move in the same direction. It's the net effect that we see. I think you're missing the fact that conventially current was thought to be the flow of positive charges.
Let's consider an example (something less complex than the diode example you've mentioned)
Consider an area element of a conductor and view it in a direction along its plane.
Let there be both positive and negative charges (yes, these are electrons (for a metallic conductor) but for the time being let them be positive and negative charges) to the left and right of the element.
If a charge +q flows from the left to right we say that a current is flowing from left to right. If it were flowing from the right to the left a current would be flowing from right to left. For a charge -q a current is said to flow from the left to right if it moves from right to left (transport of a negative charge from right to left can be visualised as a transport of an equal amount of positive charge from left to right) and similarly the other case. When we say that a current I is flowing from the left to the right it is due to a net charge Q=+q-q flowing across the area element from left to right (in the time frame we have chosen). It could be that no negative charge is flowing from the right to left. In such a case the current would be due to the positive charges solely. Similarly it could be that no positive charge flows from left to right. In this case the current would be due to negative charges flowing from right to left only. The more general case assumes the net +q-q flowing from left to right (if the current is flowing from left to right).
Now getting to the part of your question in which you address a diode being reverse biased. Who says there isn't a current when you reverse bias it? There is a current! But it's so negligibly small (in microamperes) compared to the current we get in forward bias (milliamperes) that we can neglect it and say that there is no current in reverse bias. Think about it. Both differ by a factor of 10³. Current in the reverse bias is limited by the number of minority charge carriers present in either wafers of a junction diode.
Hope that clears your doubt.
Causes of the miniscule reverse current.
(Sorry for the poor quality picture. I've cropped it from a screenshot of one of the pages of an e-book I had. Ignore the W, that's was there in another context).
You can see that the diode is reverse biased. The external biasing provided by the battery sets up an electric field which is in the same direction as that produced by the space charge regions in the depletion region.
Now any hole on the n-side (remember that both holes and electrons exist in any type (p or n) of a doped semiconductor, it's just that one is in excess) would move towards the depletion region because of the field due to the battery. If it gains enough kinetic energy to just enter the barrier region, the field there will push it to its majority charge region (I.e. the p-side).
Similar is the case with an electron on the p-side. A natural doubt would be why does that make the current small? The answer to that is the fact that the concentration of minority charge carriers (electrons in p-type and holes in n-type) is much much smaller than the concentration of majority charge carriers (holes in p-type and electrons on n-type). Since their (minority charge carriers') concentration is much much smaller their flow is much much slower.
I hope that that's a bit intuitive to you. If it isn't, here's a weird analogy: Ten people jumping on a trampoline have more chances of breaking it than a single person jumping on it. The ten people here may be thought of as those causing the large (~milliamps) current in forward bias while the single person may be thought of the person causing a negligible current (~microamps) in the reverse bias.