Is this Bleichenbacher '06 style signature forgery possible? (Or more like, why isn't it?)
Now, I'm wondering about the case where an implementation does check that the hash is right-justified, but doesn't check the previous bytes at all.


That case is completely broken for $e=3$


  Assuming that we're still talking about exponent 3, someone could imagine that there might be a way to find a perfect cube that ends with that particular hash, thereby utilizing the more significant digits that are not checked by the code. 


That actually pretty easy, with the assumption that the lsbit of the hash is a 1 (or more generally, that the number of 0's at the end of the hash is a multiple of 3); this assumption is necessary, because if it is not true, no perfect cube will exist.  It turns out if this assumption is true, then a cube will always exist.

The most straight-forward approach to find it is to do it iteratively; if we have a value $x^3 = h \pmod{2^n}$, then we can easily find a value $y^3 = h \pmod {2^{n+1}}$, as either $y = x$ or $y = x + 2^n$, and it's easy to check which one it is.  So, we just start with a small $n$, and just repeatedly add bits to $x$ until, when cubed, it generates the entire hash in its $n$ least significant bits.


  Is it possible to sign some message feasibly by somehow utilizing the structure of the particular hash in context of these (hash, perfect cube) pairs?


The above procedure works for more than half of all possible messages (assuming the hash function acts as random), hence the attacker can easily pick a message that he likes.