Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not identical to his $5$-piece dissection from unfoldings.
[![TetraCubeGood][1]][1]
The tetrahedron is unfolded 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. I consider this a solution, starting from surface unfoldings, and have accepted Gerhard's answer.