Is it possible to cover a $8 \times8$ board with $2 \times 1$ pieces? 
We have a $8\times 8$ board, colored with two colors like a typical
  chessboard. Now, we remove two squares of different colour. Is it
  possible to cover the new board with two-color pieces (i.e. domino
  pieces)?

I think we can, as after the removal of the two squares, we are left with $64-2=62$ squares with $31$ squares of each colour, and - since the domino piece covers two colours - we can cover the new board with domino pieces.
But how should one justify it mathematically?

Yes.
Consider the board initially
covered with dominoes.
After the two squares
have been removed,
the board has two holes.
If the two holes
were in the same row,
slide the dominoes
between the holes
until one of the holes
is filled.
This will leave
two adjacent holes
that can be covered
by a domino.
If the holes
are in different rows,
slide dominoes 
from the lower row
to the upper row
up
until one of the holes is filled.
That leaves the lower row
with two holes
which can be filled
as described above.
If the two rows are adjacent,
slide the upper row
left or right
until its hole is filled
and the hole has moved
above the hole
in the next row.
This can be filled
by a domino.
