To determine who has a winning strategy, we need to analyze the final position of both players after $n$ moves. The key observation is that the net rotation of both players depends on the parity of $n$.

1. Understanding the Moves:
- At odd times $2k-1$, player $A$ can either stay stationary or turn $90^\circ$ clockwise, and player $B$ follows.
- At even times $2k$, player $B$ can either stay stationary or turn $90^\circ$ clockwise, and player $A$ follows.

2. Net Rotation Analysis:
- Each pair of moves (one by $A$ and one by $B$) results in a net rotation of $0^\circ$ or $90^\circ$ clockwise.
- If $A$ and $B$ both choose to turn $90^\circ$ clockwise, the net rotation is $180^\circ$.
- If one chooses to turn and the other stays, the net rotation is $90^\circ$.
- If both stay, the net rotation is $0^\circ$.

3. Modulo 4 Analysis:
- The direction faced by the players after $n$ moves depends on $n \mod 4$:
- $n \equiv 0 \pmod{4}$: Final direction is North.
- $n \equiv 1 \pmod{4}$: Final direction is East.
- $n \equiv 2 \pmod{4}$: Final direction is South.
- $n \equiv 3 \pmod{4}$: Final direction is West.

4. **Case (a): $n = 2013^{2012}$**
- Calculate $2013^{2012} \mod 4$:
$$2013 \equiv 1 \pmod{4} \implies 2013^{2012} \equiv 1^{2012} \equiv 1 \pmod{4}$$
- Since $2013^{2012} \equiv 1 \pmod{4}$, the final direction is East.
- Therefore, $A$ wins.

5. **Case (b): $n = 2013^{2013}$**
- Calculate $2013^{2013} \mod 4$:
$$2013 \equiv 1 \pmod{4} \implies 2013^{2013} \equiv 1^{2013} \equiv 1 \pmod{4}$$
- Since $2013^{2013} \equiv 1 \pmod{4}$, the final direction is East.
- Therefore, $A$ wins.

The final answer is $\boxed{ A }$ wins in both cases.