In order to define integration of differential forms on a manifold $M$, one must choose an orientation of $M$.
Strictly speaking, this means there isn't enough information to answer your question, but people often don't
explicitly specify the orientation, and instead use whatever makes the most sense. The point is, I'm going to
assume the standard orientations of $S_X^1\times S_Y^1$ and $S_Y^1\times S_X^1$, in which case they're oppositely
oriented and you're correct that $[S_X^1\times S_Y^1] = -[S_Y^1\times S_X^1]\in H_2(T^2)$. But if you had a
different orientation in mind, the answer could be different.
Suppose $V$ and $W$ are oriented vector spaces, that $(e_1,\dotsc,e_m)$ is a positively oriented basis of $V$, and
that $(f_1,\dotsc,f_n)$ is a positively oriented basis of $W$. The standard orientation on $V\oplus W$ is the one for
which the basis $(e_1,\dotsc,e_m,f_1,\dotsc,f_m)$ is positively oriented. This extends to the orientation
convention for products of manifolds: if $M = N_1\times N_2$, then at any $(x_1,x_2)\in M$, $T_{(x_1,x_2)}M =
T_{x_1}N_1\oplus T_{x_2}N_2$, so the convention defines an orientation $T_{(x_1,x_2)}M$ given orientations of
$T_{x_1}N_1$ and $T_{x_2}N_2$.
The point is that the orientations on $N_1\times N_2$ and $N_2\times N_1$ may differ, and they do in this
situation. The orientation on $S_X^1\times S_Y^1$ is the one in which the basis $(\partial_x, \partial_y)$ of
$T_{(x,y)}T^2$ is positively oriented, and the orientation on $S_Y^1\times S_X^1$ is the one in which $(\partial_y,
\partial_x)$ is positively oriented. The change-of-basis matrix between these bases is $\begin{pmatrix}0 & 1\\1 &
0\end{pmatrix}$, which has negative determinant, so these two orientations are opposite. Therefore the answer to
your question is no; in fact,
$$\int_{S_X^1\times S_Y^1} \mathrm dX\wedge\mathrm dY = -\int_{S_Y^1\times S_X^1} \mathrm dX\wedge\mathrm dY.$$