Indeed, if $\sigma(X_1) = \sigma(X_2)$ then $E[Y | X_1] = E[Y | X_2]$ almost surely, thus $E[Y | X_1] = u_1(X_1)$ and $E[Y | X_2]=u_2(X_2)$ for some measurable functions $u_1$ and $u_2$ such that $u_1(X_1)=u_2(X_2)$ almost surely.
One can then define $E[Y | X_1=x]$ as $u_1(x)$ and $E[Y | X_2=x]$ as $u_2(x)$ for every $x$ in the target set of $X_1$ and $X_2$. This does not entail that $E[Y | X_1=x]$ and $E[Y | X_2=x]$ coincide since, in general, $u_1(x)\ne u_2(x)$.
To sum up, the condition that [$u_1(X_1)=u_2(X_2)$ almost surely] does not imply that [$u_1=u_2$].
Edit: Perhaps a simple example can help. Assume that $Y=6X_1=3X_2$, hence $X_2=2X_1$. Then $$E[Y | X_1]=E[Y | X_2]=Y=6X_1=3X_2,$$ hence, for every $x$, $E[Y | X_1=x]=6x$ and $E[Y | X_2=x]=3x$. If one selects some $\omega$ in $\Omega$ and one measures $X_1(\omega)=x_1$ and $X_2(\omega)=x_2$ then $x_2=2x_1$ hence $E[Y | X_1=x_1]=6x_1$ and $E[Y | X_2=x_2]=3x_2$, which implies that $$E[Y | X_1=x_1]=E[Y | X_2=x_2].$$
More generally, if $\sigma(X_1)=\sigma(X_2)$, there exists some invertible bimeasurable $v$ such that $X_2=v(X_1)$ almost surely hence
E[Y\mid X_2]=u_2(X_2)=u_2\circ v(X_1)=E[Y\mid X_1],
thus, for almost every $x_1$ (with respect to the distribution of $X_1$),
E[Y\mid X_2=v(x_1)]=u_2\circ v(x_1)=E[Y\mid X_1=x_1].