It seems to me the answer should be yes, and is certainly yes in case $f$ is smooth. I'll sketch an argument in the smooth case; I think the details can probably be filled in more generally. (Note that if you just assume $X$ is nonsingular and the characteristic is 0, you can easily reduce to the smooth case by generic smoothness)

Put

$Z=\{ (x_1,x_2): f(x_1)=f(x_2)=y \textrm{ and } x_1,x_2 \textrm{ lie on the same component of } f^{-1}(y)\} \subset X \times_Y X$.

Check that $Z$ is a subvariety of $X\times_Y X$ with the same dimension as $X\times_Y X$ (this is easy if $f$ is smooth, and I think it should be true generally; in the general case, it is probably best to throw out some of the "bad" fibers of $f$ and take a closure to define $Z$). Then since $X\times_Y X$ is irreducible, we have $Z = X\times_Y X$, which gives the desired conclusion.