The answer is no. Let's color the $8\times 8$ board in the following way:

Then, each $4\times 1$ piece will cover equal number of black and white squares. But a $2\times 2$ piece will cover one square of one color and three squares of the other color. It follows that fifteen $4 \times 1$ pieces and one $2\times 2$ piece cannot cover equal number of white and black squares. However, the numbers of black and white squares on this board are the same. Therefore, we conclude it is impossible to do so.