Would there be a difference if deuterium is embedded instead of protium (regular hydrogen) in acids?
Yes, there would be a difference. The acids would be weaker.  Why? Because the heavier isotope lowers the zero-point energy (the lowest possible energy) in the acid molecule, making the acid's bond to the D stronger than it would have been to the H. Acids in $\rm D_2O$ are weaker than they are in ordinary water, too.

Any reaction that involves dissociation of these acids will run more slowly, too. This is called "the kinetic isotope effect". You can read more about that here: http://en.wikipedia.org/wiki/Kinetic_isotope_effect

You might want to ask further questions: why is the zero point energy lower for a bond to D than a bond to H? How exactly does the lower zero point energy make the bond stronger? How exactly does it affect the rates of reactions that include acid dissociation?

Ask away!

If you'd like to see how substituting a D for an H in acids affects the acid dissociation constant for particular acids, here's a good reference: 

R. A. Robinson, M. Paabo, R. G. Bates, "Deuterium Isotope Effect on the Dissociation of Weak Acids in Water and Deuterium Oxide", J. Res. Nat. Bureau. Standards, 73A(3), 1969, pp 299-308.