Again, to make it clear,
1. If we are already given that the input relation is equivalence relation, then the strongly connected component would give the correct partition.. we can find using DFS numbering method - i think it has complexity O(V+E) for adjacency lists, and O(V2) for adjacency matrix, in any case less efficient than the disjoint set DS.
2. but for general case of a relation, this might not be true..
Strongly connected component need not necessarily imply partitions i think..
Take this graph for example..