Another reason it doesn't much matter is that, to make things more consistent, we're changing what it means in Perl 6. The notation no longer has anything to do with how the arguments are processed--there are now flattening operators for that. Instead, when you say &foo, you're explicitly marking "foo" as a noun rather than a verb. As a noun, it always returns a scalar reference in scalar context, and like any other reference, it is not dereferenced unless you do so explicitly. In the case of &foo, it is not called unless you say &foo() or some such.

Interestingly, this completely removes the Perl 5 distinction between foo(1,2,3) and &foo(1,2,3). Those are handled identically in Perl 6. What changes is the meaning of bare &foo. You now always have to pass @_ explicitly if that's what you mean.