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No, there is already a diode to protect against reverse polarity but current is limited in this case by a 47ohm resistor. If the power is connected with reverse polarity, the current through that resistor will be about 200mA. That means the resistor will have to dissipate around 2 watts.

So my original question - is it better to use a 1/4 watt resistor that will act as a fuse, or up the rating to 2 watts?   

I meant that the 47ohm would need to be 2W to not burn up if reverse polarity power is accidentally connected. In normal use 1/4W would be more than enough.

I'm about to build a Faultline and, as the pdf doesn't specify power ratings, I've chosen 1/4W resistors. I'm sure they'll be fine but I wonder about the 47ohm resistor in series with the 9V power coming into the PCB. I guess this is to limit current through the polarity protection diode. To be safe, the 47ohm would need to be rated at around 2W - or is it better to use a lower rating and have the resistor function as a fuse?