The forward voltage of that diode @ 50A is roughly 0.8V; the power lost is thus 0.8V*50A=40W. It’s a dual device, so if you’ve got 50A running through both, the figure would be double. If you’ve ever grabbed a 40W incandescent lamp that’s been running a while, you can appreciate that it’s a fair amount of heat to get rid of. So yes, a heat sink is in order.
This post describes the process of heat sink selection. Assuming that you’ve got a 40W thermal load, a heat sink with an effective thermal resistance of about 2°C/W or less would probably be a reasonable choice.