There are far too many unspecified variables to give a definitive answer to this question. A straightforward interpretation, and the one answered here, is of heating an object of these two colors by shining light onto it. Even with this assumption, the question cannot be answered with certainty, but by making a few additional assumptions (such as, an equal distribution of photons (basic units that makes up all light) across the colors in the light and the same type of material), the green colored object will probably get hotter. There are many details which would change this though. Some of these are covered here, and other possibilities are given at the end of this response.
In this case of heating an object by light, all of the energy increasing the temperature of the object is assumed to come from the light. The object which absorbs more of the light will get hotter. The energy in that light depends on the wavelength(s) and the intensity of that light.
For this problem, consider the energy of light to be carried in discrete packets called photons. The energy (Ep) of a single photon of light with a particular wavelength is given by Ep = h*c/L, where h = 6.62607*10-34 m2-kg/s is a number called Planck's constant, c = 2.998*108 m/s, and L is the wavelength of the light. The total energy (E) is then given simply by the energy per photon multiplied by the number of photons absorbed, that is E = Ep * n where n is the number of photons. A "more intense" light will have more photons coming in per unit of time.
The color of an object depends on the wavelength(s) of light reflected (i.e., not absorbed) by the object, and there are two possibilities.
If an object reflects only light with wavelengths in the "red" portion of the visible light spectrum, that object will appear red because that is the only wavelength reaching the eye; light at all other wavelengths is absorbed. Alternatively, an object could reflect all wavelengths except those of a certain color. In this case, the object would appear to be the complement (sort of the opposite) of the absorbed color. For example, absorbing only "red" will make an object appear blue-green.
An object which absorbs all wavelengths of visible light will appear black, and one which absorbs none will appear white.
Relating this now to the question of whether a pink-colored or red-colored object will get hotter, what one needs to determine is essentially which object can absorb more light. Doing this requires many assumptions. For now, take that
(1) the colors are caused by reflecting some red and violet light (reflecting ~90% of the red and 40% of the violet results in pink ) and (essentially all) green light, respectively;
(2) the light has photons equally distributed across all and only visible-light colors (that is, there are as many "red" photons as there are blue, yellow, etc., and there is no light with wavelengths outside of the visible range);
(3) the objects absorb and reflect their respective wavelengths completely;
(4) apart from color, the materials are the same.
Using these, the energy reflected by the pink object can be estimated as (fraction red reflected * energy of red) + (fraction of violet reflected * energy of violet) = 0.9*1.8 eV + 0.4*3.1 eV = 2.86 eV.
Similarly, the energy reflected by the green object is 1*2.3 eV = 2.3 eV. (Energy per color here ; eV is a unit of energy.)
Because the pink object reflects more energy, it must absorb less, which means that less of the light's energy is being used to heat the object. Thus, the green object would get hotter.
[Some responses on Scienceline deal with - other topics - related to color and heat .]
As stated at the beginning, reaching this result required many assumptions. If the light has a different intensity distribution (for example, sunlight has nearly uniform power [power = energy per time] in the visible range rather than uniform number of photons), then the energy reflected would change. If, unlike this example, the materials are different, then they could have different surface finishes (i.e., one shinier than the other) which would affect the reflected energy , and they could have different heat capacities , which is a measure of how much energy is needed to change the temperature of an object.
One could also (try) to consider absorption in the case of coloring due to the complement effect described above. As the temperatures of the objects increase above that of their surroundings, then the issue of emitting (the opposite of absorbing) heat comes into play, as one object/material might be a better emitter than the other. One especially major assumption is that the heating is caused by absorption of light while emission is ignored. This means that the objects can heat up but not cool off. Conveniently, good absorbers of heat tend to also be good emitters, so some of the above can be extended fairly straightforwardly. If the heating were accomplished by some method other than light, then heat emission as well as a host of other factors affecting the heat transfer from the hot thing to the pink/green objects would have to be considered.