You could create a dict comprehension of just the elements whose values are None, and then update back into the original:
tmp = dict((k,"") for k,v in mydict.iteritems() if v is None)
mydict.update(tmp)
Update - did some performance tests
Well, after trying dicts of from 100 to 10,000 items, with varying percentage of None values, the performance of Alex's solution is across-the-board about twice as fast as this solution.