In general, degeneracy in quantum mechanics means that there are at least two states that have indistinguishable energy. In the case of a degenerate photon pair, the two photons have will the same frequency (as E = $\hbar$ $\omega$).
This is, of course, oversimplifying things slightly. The above describes the case when the photons are each in a single mode. In reality, most photon sources will produce photons which are multimode. But we won't go there!
It's hard to elaborate further without the specific context in which "non degenerate" appears. I'd imagine that in the context of photon pairs it is in relation to quantum inference. In the case of the paper discussed in a previous answer (arxiv.org/pdf/1304.1490.pdf) this is indeed the case.
I won't talk about quantum interference here, but maybe discussing how the states are labeled in the above paper will clear up what they mean by degenerate and non-degenerate.
Degenerate photons pairs- both in the same mode- are given by states like
$\left|n_{A} n_{B}\right>$
$n_{A}$ and $n_{A}$ are simply the number of photons in the output ports A and B of their experiment. As the photons are degenerate, there is no need to give a label to show which photons have which frequency. They are all the same!
For non degenerate states, the photons have frequencies labeled by i and s. Non degenerate pairs exist in states like
$\left|n_{i} n_{s}\right>_A$ $\left|n_{i} n_{s}\right>_B$
A and B still label the output ports, but we now have to label which mode the photons are in as well (i or s). For example $\left|1_{i} 0_{s}\right>_A$ $\left|0_{i} 1_{s}\right>_B$ is a state where there is one photon of mode i in port A and one photon in mode s in port B.
I hope this helps. If you're completely new to quantum optics, then a good place to start might be "Introduction to Quantum Optics: From the Semi-classical Approach to Quantized Light" by Grynberg, Aspect and Fabre.