First of all, note that if $(3,4)$ is disallowed, by symmetry, $(4,3)$ is disallowed as well. By building the symmetric closure of the forbidden set, we find that the following pairs are disallowed: $(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4)$.
Next, by reflexivity, every equivalence relation must contain $(a,a)$ for $a = 1,\ldots, 6$.
Furthermore, if $(1,5)$ and $(5,2)$ were part of the relation, then by transitivity, $(1,2)$ would be part as well, which is impossible. Thus, there are at most one $a$ and one $b$ such that $(a,5)$ and $(b,6)$ (as well as $(5,a)$ and $(6,b)$ by symmetry) are part of the relation.
If $(5,6)$ is part of the relation, we cannot have $(a,5)$ and $(b,6)$ in the relation for $1 \le a,b\le 4$, $a \neq b$ by another transitivity argument.
Therefore, there are two classes of equivalence relations: Those containing $(5,6)$ and those not containing $(5,6)$.
With further symmetry and transitivity arguments we find that if $(a,5)$ and $(5,6)$ are in the relation, so is $(a,6)$ (and all their symmetric forms), and vice-versa. Therefore, there are four equivalence relations with $(5,6)$ and $(a,5)$ for $1 \le a \le 4$. Add the case where none of these pairs is added, and you find there are exactly 5 cases with $(5,6)$.
In the case that $(5,6)$ is not in the set, one finds that there are three main sub-classes: The identity relation (one member), sets that contain $(a,5)$ (four members), sets that contain $(a,6)$ (four members) and sets that contain $(a,5)$ and $(b,6)$ for $a\neq b$ ($4 \cdot 3 = 12$ members). All in all, there are 26 different equivalence relations.