Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not exactly what he had in mind.
[![TetraCubeGood][1]][1]
The tetrahedron is unfolded and reshaped to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.
This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles, and two more pieces to cut out the right triangle to reshape the rectangle back to the tetrahedron parallelogram. $7$ pieces altogether. I consider this a solution, and have accepted Gerhard's answer.