My thoughts...
A volume of oil trapped will absorb little or no shock. The bulk modulus of oil is far less than that of air, it will compress slightly, however, at such low pressures the compression will be almost zero.
Energy that is stored in oil will be transfered to heat if it is not used. When compressing the oil, if it is done isothermally, i,e slowly, there will be almost no heat increase. When pressurised 120 time per second, the oil may tend to warm up. However, the amount of energy required to compress the oil could be very low. By the way, what is the volume of the actuator? The size of the actuator makes a big difference to the oil temperature. Specific heat capacity of oil...the more oil the more energy required to raise the temperature. The bigger the surface area of the actuator, the more heat it will disipate.
If the "other" end of the actuator is to be charged by a pump you may find that there is an inherent ripple that is transmitted from the pump. Almost all hydraulic pumps have a ripple of some sort. In most hydraulic systems the ripple is absorbed by the components. In this application there would be no way to absorb the ripple. Depending on the rotational speed and the type of pump, the frequncy of the ripple could be as high a 500Hz. If the pump is working upto 1200 PSI, that is the pressure that will be seen in the actuator.
As I amagine the system to work, there will no oil flowing at the actuator and the volume of oil is such the low power required to raise the pressure will limit the heat input.
There are many other hidden factors that can effect that heat input, can you provide more details about the pump and the pressure controls etc...?
Regards
Adrian;-)