It will only impact your power dissipation. Precision tolerances are harder to find for larger parts, but I imagine you are just doing 1% which is easy.
50mA through a 100R resistor = 50^2*100 = 250mW.
So 1/4W or 1W would be enough, but if you go up to 100mA you would need the 1W.
You can use a resistor rated for a higher wattage at the same resistance to replace a lower wattage resistor.
Caveat Emptor (Buyer Beware) - when going to a higher wattage resistor (5 Watts and over), be aware that most higher wattage resistors are wire wound. Wire wound resistors have something extra - inductance. The inductance can cause problems in certain types of circuits. The inductance of the wire wound resistors is not usually specified.
Calculating the resistor wattage that you need:
P=E^2/R, or I^2 x R, or E x I. Then DOUBLE the power requirement, and use a wattage at least as large as the result.
Skeebopstop gave you:
50mA through 100R = 50^2*100 = 250mW
But then you need to double the 250mW to 500mW, or 1/2 Watt.
A 1/4W resistor would work, but it would be operating at it's maximum rating and would have a very short life.
I don't know how many mAmps 3 double A batteries provide. The schematic called for a 1/4W resistor but I only have 1W resistors on hand with the same resistance. Would it be fine to use the !W or the batteries will get drained faster with the 1W resistors.
You can always substitute a different component that either has the exact wattage ratings or more than the wattage ratings. For example, if I need a 1/4W resistor, I could actually use a 10W resistor as long as it has the same resistance value. Moreover, you can't go under the listed wattage rating, because the component might overheat and damage that particular part, in addition to other parts in certain circumstances.