For "no evidence of X" to be no evidence against X, it would need to be true that you're no more likely to see evidence of X if X than if not-X. That's a pretty weird situation. It might happen if X were totally unobservable, or if you were being systematically deceived. (I would then say that nothing is evidence of X.) But the usual case is that evidence of X is more likely when X than when not-X, and then absence of evidence of X is evidence of not-X.
More formally: I take "A is evidence for B" to mean "Pr(B|A) > Pr(B)". "Absence of evidence is evidence of absence" means that if Pr(B|A) > Pr(B) then Pr(~B|~A) > Pr(~B). Assuming that none of the events involved is outright impossible, we can equivalently write the condition for being evidence as Pr(B|A) > Pr(B|~A), and the question is whether this implies Pr(~B|~A) > Pr(~B|A).
Well, we can express everything here in terms of four probabilities which I'll call AB, Ab, aB, ab. "Ab" means Pr(A & ~B), etc.: capital for true, lowercase for false. Now Pr(B|A) > Pr(B|~A) is AB/(AB+Ab) > aB/(aB+ab), and Pr(~B|~A) > Pr(~B|A) is ab/(aB+ab) > Ab/(AB+Ab).
The first, on clearing denominators, becomes AB(aB+ab) > aB(AB+Ab); a term AB.aB on each side cancels, leaving AB.ab > aB.Ab.
The second, on clearing denominators, becomes ab(AB+Ab) > Ab(aB+ab); a term ab.Ab on each side cancels, leaving ab.AB > Ab.aB.
And, lo and behold, these two conditions are the same. In other words, barring pathological cases where some of the probabilities are zero, "A is evidence for B" is exactly equivalent to "Not-A is evidence for not-B".
But the strength of the evidence can be very different in the two cases, as in Hempel's paradox: let A be "the object I just observed is a white crow" and B be "crows are not all black"; clearly A is very good evidence for B; so not-A is evidence for not-B; but if what you just observed is (say) a red car then it's very very weak evidence that all crows are black. But it's evidence, none the less: you just eliminated one possibility for something that might have been a counterexample.