Cyclopropanone is a reactive species, as such there aren't many experimental reports in which it is characterised, however some (possibly questionable data) does exist:
Cyclopropanone is a highly reactive compound. It is stable for a few days at liquid nitrogen temperature but rapid polymerisation takes place above o or even below if traces of water are present. The reaction is highly exothermic.
Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.19660851113
The NMR spectrum (neat) shows a singlet at 1.72 p.p.m. relative to tetramethylsilane.
Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.196608511
It appears that in the absence of a chiral environment, all of the protons are in a single environment at 1.72 ppm (Hans Reich quotes a single peak at 1.65 ppm) which is what we'd expect (the molecule is essentially planar, and all of the protons are equivalent).
As you point out in your question, replacing any one of the protons (hypothetically) would yield two enantiomers, which, in the presence of a chiral environment, should be distinguishable. It can therefore be concluded that in a chiral environment, the protons should appear to be enantiotopic.*
* there are two slight caveats to this. Firstly, the compound would still be unstable and prone to polymerising, therefore there is no way of actually testing this. Secondly, the chiral environment must be able to interact with the molecule, for things like chiral alcohols, lanthanide shift reagents are (/used to be) commonly used, but I'm less convinced how you put a ketone into a chiral environment.