Over the reals one has a parametrization of the rotations given by
$$
\begin{pmatrix}
\cos t & - \sin t \\
\sin t & \cos t
\end{pmatrix}
$$
In other words
$$
\begin{pmatrix}
x & - y \\
y & x
\end{pmatrix}
$$
with $x^2 + y^2 =1$ with reals $x,y$.
Now if one wishes to restrict to the rationals one asks for rational solutions of
$x^2 + y^2 =1$ (every such solution will give a rotation and one must not have any irrational entries in the matrix; also cf. André Henriques answer).
There is a well-know rational parametrization of these solutions given by
$$
x(t) = \frac{1-t^2}{1+t^2} \, , \, y(t) = \frac{2t}{1+t^2}
$$
with $t$ rational and the one additional solution $(-1, 0)$.
(The geometric idea is that the $t$ is the slope of a line through $(-1, 0)$ and the respective solution the other intersection point of this line with the circle.)
So one would get all the rotations as
$$
\begin{pmatrix}
\frac{1-t^2}{1+t^2} & - \frac{2t}{1+t^2} \\
\frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2}
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
-1 & 0\\
0 & -1
\end{pmatrix}
$$
This is also closey related to parametrization of Phytagorean Pythagorean triples.
