Solution:
Answer: 5.

Tetrahedral faces are triangular, so each cube face requires at least two tetrahedral faces. So at least 12 tetrahedral faces are needed in all. At most three faces of a tetrahedron can be mutually orthogonal (and no two faces can be parallel), so at most 3 faces from each tetrahedron can contribute towards these 12. So we require at least 4 tetrahedra to provide the cube faces. But these tetrahedra each have volume at most $1/6$ ($1/3$ x face area x 1, and face area is at most $1/2$). So if we have only 4 tetrahedra in total then their total volume is less than the cube's volume. Contradiction. Hence we need at least 5 tetrahedra.

It can be done with 5: lop off 4 non-adjacent corners to leave a tetrahedron. More precisely, take the cube as $ABCD A'B'C'D'$ with $ABCD$ horizontal, $A'$ directly under $A$, $B'$ directly under $B$ and so on. Then the five tetrahedra are $AA'BD$, $CC'BC$, $DD'AC'$, $BB'AC'$, $BDA'C'$.