Can two rational rotations $F_2 = \langle A, B \rangle \to SO(3)$ efficiently approximate the $3 \times 3$ identity matrix?

Setting $a_0=A^7, b_0=B^7$ and then $a_{n+1}=[b_n^{-1},a_n], b_{n+1}=[a_n,b_n]$ seems very efficient. The length grows like $C \alpha^n$ with $\alpha= \frac{3 + \sqrt{17}}2$ and the operator norm distance to the identity matrix gets squared in each step, since $$\|1 - [a,b]\| \leq 2 \|1-a\| \cdot \|1-b\|.$$ I used a similar construction in On the length of the shortest non-trivial element in the derived and the lower central series. I do not think this is best possible, but it is very explicit.

Have you tried b^{-34}*a^{34}

It gets pretty close...
$b^{-34}*a^{34} = \left[ \begin{matrix} 0.9937 & 0.1119 & 0 \\ -0.1112 & 0.9875 & -0.1119 \\ -0.0125 & 0.1112 & 0.9937 \end{matrix} \right]$

in fact chaining b^-17*a^17*...*b^-17*a^17 an even number of times will get you closer while maintaining unit determinant...

$b^{-17}a^{17}b^{-17}a^{17}b^{-17}a^{17}b^{-17}a^{17}=$ $\left[ \begin{matrix} 1.0000 & 0.0002 & -0.0063 \\ -0.0002 & 1.0000 &-0.0002 \\ 0.0063 & 0.0002 & 1.0000 \end{matrix} \right]$

Are these results product of rounding errors?

No!

The geometric meaning is that $17 \cdot \arccos(3/5) \approx 5\pi$, so $a^{17}$ is like rotating two and a half times around the $\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right]$ axis, and $a^{34}$ is 5 way round, so you obtain the identity (approximately).

You could chase down the decimals after $\arccos(3/5)$ and get so close to identity as you wish. And the same goes with b. The question about combinations of powers of multiples other than 17 is far more difficult.