Hydrogen and deuterium exchange
If you are interested in making deutero-chloroform, there are better ways to do it.  

From a purely mechanistic standpoint, chloroform isn't a very strong acid ($\ce{pKa \sim 15}$) so $\ce{D2O}$ is unlikely to deprotonate it at a useful rate.  Further, chloroform isn't very soluble in water making your desired reaction even less likely (but you could engineer your way around this problem by using a phase transfer catalyst).  

Your suggestion to use a stronger base is reasonable, but the problem you run into here is that a different reaction is brought into play.  Specifically, once a stronger base like potassium t-butoxide (in t-butanol) is used, dichlorocarbene formation occurs.  Once the trichloromethylcarbanion is formed, it quickly ejects $\ce{Cl^-}$ to form dichlorocarbene.  Dichlorocarbene is an extremely energetic species and would quickly react with whatever else is around - perhaps the t-butanol to produce dichloromethyl-t-butyl ether.