For any surface, the absorption and emission coefficient is the same at a given wavelength. You didn't specify whether your two surfaces were "equally grey" (that is, the reflectivity was constant for all wavelengths, just a different value) or "colored" (that is, one might be blue, and have low emissivity in the IR, while the other is white).
I believe that difference matters. In atmospheric science, we recognize albedo (the fraction of incident light reflected) and emissivity (the fraction of black body radiation emitted). When you have zero albedo and 100% emissivity, the equilibrium temperature of the object will be that of a black body in space. But if the albedo is high the object will be cooler than that; and if the emissivity is low, the object may be hotter. The equation (see this answer to see the derivation) is
$$T_o = T_S \sqrt{ \frac{ R_S \sqrt{\frac{1-\alpha}{\overline{\epsilon}}} }{2 D_{o-s} } }$$
Where I use the subscript $_o$ for the object and $_s$ for the sun.
$R_s$ is the radius of the sun, $D$ is the distance from the (center of the) sun, $\alpha$ is the albedo (properly averaged over the wavelengths of incident light), $\epsilon$ the emissivity (again, properly averaged over the wavelengths of the emitted light - by the Wien displacement law we know the emitted wavelength will be longer because the object is cooler than the sun; so in general $(1-\alpha) \ne \epsilon$ and the object's equilibrium temperature depends on both of these.
Now if the object is "uniformly gray" (whether $(1-\alpha)=\epsilon=0.9$ or $(1-\alpha)=\epsilon=0.1$), the result will be the same. So a black body, a grey body, and a white body, will all reach the same equilibrium temperature. But a "blue body" might not.
