Not necessarily. Unless the two transistors are perfectly matched (highly unlikely), one of the pair will end up conducting more than the other. For low power applications like driving a 1W led for example, this wouldn't be an issue as you would be operating a long way from the transistors rated current limit. However in high power applications such as this one, a lot more care is needed.
For example (and this is just a hypothetical, don't take these figures as accurate):
Say you pump 90A through the board. You have two transistor rated at 45A MAX. Transistor A has an on resistance of 0.01 Ohm, transistor B an On resistance of 0.015 Ohm.
You would have essentially a 0.006 Ohm resistor (the two in parallel), which at 90A would drop 0.54V.
Consider transistor A. You would have V/R = I = 0.54/0.01 = 54A. BOOM, that transistor overheats and blows. Then you just have transistor B. That would now take the full 90A, and BOOM it too blows.
As you can see, in this hypothetical situation, even a difference of 5 thousandths of an ohm difference is On resistance would be enough to destroy both FETs.
You would need to replace the FET with one which has a higher current rating, but with similar gate characteristics to ensure compatibility. Something like this might be a possibility: