Just to mark this question as answered, the comment by Tobias Fritz is spot on. In <a href="https://math.stackexchange.com/a/2122958/79593">this answer</a> to a Math Stack Exchange question, Francisco Santos completely resolves your questions. On the one hand, the answer to Question I is no: combinatorially equivalent polytopes can have different triangulations. On the other hand, the answer to Question II is: yes, there is at least one triangulation they all share (the so-called "pulling triangulations").