Here is a dissection based on Gerhard's latest idea. I don't add this directly to his post because I am not certain it is 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.
Indeed this appears to be $5$ pieces, or $6$ to reshape the final rectangle back to tetrahedron parallelogram.