This should work.

1. The colouring:
>! [![edge-coloruing][1]][1]

2. The sequence:
>! R, R, R, R, R, R, B, B, B, B, B, i.e., 6 R's followed by 5 B's.

Bonus:
>! Label the cities 0, …, 10 starting from Rome and going in anticlockwise direction. Then, starting from city $i$, the number of clockwise steps ($c_i$) must be $i$ more than the number of anticlockwise steps ($a_i$) *modulo 11*, i.e., $c_i - a_i = i \mod 11$. But we want the total number of steps to be the same starting from any city. In particular, $c_0 + a_0 = c_1 + a_1$. If $c_0 = a_0$, then $c_1 + a_1$ must be even, but then $c_1 - a_1 - 1$ must be an odd multiple of 11, which means $c_1 + a_1$ is at least 12 which is more than the above solution. Therefore, $c_0 - a_0$ is a nonzero multiple of 11 and the above solution is minimal.

  [1]: https://i.sstatic.net/v8kVJ2Ko.png