Solution: No.

Any $2 \times 1$ or $1 \times 2$ block always covers one dark square and one light square. Thus any configuration of these blocks covers an equal number of light and dark squares.

In board $B$, there are $30$ dark squares and $32$ light squares. Assume that there is a configuration of blocks that covers board $B$: this configuration of blocks covers a different number of light and dark squares. A contradiction.

This is known as the mutilated chessboard problem and belongs to combinatorics, a field of math that typically studies various counts and configurations of finite structures.