Short answer: diode current rating must be equal to current flowing in the load.
Long answer: The "back EMF spike" (I somewhat hate the term, since it really is just a characteristic of the inductance) results from the basic equation of the inductor, voltage is equal to inductance times the rate of change of the current (U=L*di/dt). Now, if you abruptly just stop the current, then the resulting voltage is large. Resulting voltage polarity becomes so that it resists the change in the current. Current tries to flow same direction after the disconnection of the supply. Thus the diode direction must be so that current can flow through it. As it happens, it settles reverse biased across the load to satisfy this condition.
BTW, this "back EMF" is perfectly analogous with what will happen if you short-circuit a fully charged capacitor (I=C*du/dt). Just current and voltage change roles (big current spike when voltage changes fast). It is kinda strange that shorted capacitor case is easier for people to understand.
But if you can arrange alternative path for the current so that it decays slowly to zero, then there is no spike. By connecting diode across the inductive load, a controlled path is provided for the current, current decay is slower and resulting "back EMF spike" is small. This causes same current going through the diode, what was going from outside supply before the turn-off. Thus diode must take initially same current what load takes, decaying slowly to zero.
Regards,
Janne