Can Nondet Rabin Tree automaton be determinized?
The answer is no, nondeterministic Rabin automata are more expressive than deterministic ones.

As an example, consider the language of $\{a,b\}$-labeled binary trees that contain at least one $b$.

It's easy to construct a two-state B&uuml;chi tree automaton for this language (and hence a Rabin automaton), but there is no deterministic equivalent. 

You can look at Lemma 6.1 here, for a start.