Answer: Yes, you can always. The universal strategy is to encode the secret slot with XOR over $0,\dots,63$ indices.
1) Board Coding
- Number the 64 squares as $0,1,\dots,63$.
- Defines one bit per square: heads $=1$, tails $=0$.
- Let $T$ be the secret index marked by the guard.
Calculate the XOR of all face squares:
$$
X=\bigoplus_{i:\,b_i=1} i.
$$
2) Which coin to flip
The first prisoner calculates:
$$
F=X\oplus T.
$$
- If $F=0$, no need to flip any (allowed: at most one).
- If $F\neq0$, exactly flip the coin from square $F$.
3) Correction test
Flipping $F$ changes the global XOR on $\oplus F$, then the second prisoner sees:
$$
X'=X\oplus F.
$$
Replacing:
$$
X'=X\oplus(X\oplus T)=(X\oplus X)\oplus T=T.
$$
Therefore, the final XOR read by the second prisoner is exactly the secret index.
4) Complete example
If the faces are in $\{2,5,9,12\}$, then:
$$
X=2\oplus5\oplus9\oplus12=2.
$$
If the guard dials $T=11$:
$$
F=2\oplus11=9.
$$
Square 9 is flipped and the final XOR becomes 11, which identifies the secret square.
5) Methodological readingThe trick does not depend on the initial state: the first prisoner sends 6 bits of information (index 0..63) through a single controlled inversion. This guarantees success in all cases.