Let the pieces of cardboard be placed on the grid as required. Since each piece covers exactly 4 unit squares and no two pieces touch, there is at least a 1-unit wide space between every two pieces. Therefore, if we draw a half-unit wide "no-go zone" around each piece, the areas covered by the pieces and their no-go zones will have dimensions of $2 \times 5$ and will not overlap. Since these areas extend over the edges of the grid by at most half a unit, all these areas can fit within a $11 \times 11$ square. Therefore, no more than $\lfloor \frac{11 \cdot 11}{2 \cdot 5} \rfloor = 12$ pieces can be placed on the grid.
This limit case is achievable, as shown in Fig. 37.

Fig. 37

Divide the $10 \times 10$ grid into squares of dimensions $2 \times 2$ (Fig. 38).

Note that at most 2 unit squares can be covered in any $2 \times 2$ square because otherwise at least two different pieces must be used which would then touch each other. Since there are 25 squares of $2 \times 2$, the pieces can cover at most $25 \cdot 2 = 50$ unit squares, and thus there can be at most $\lfloor (50/2) \rfloor = 12$ pieces. The fact that 12 pieces are possible is shown in Fig. 37.

Fig. 38