The answer is no, even for $d=1$. Note that for curves over a field $F$ (edit: geometrically connected ones, so that global sections of the curve are just $X$), incompressibility is equivalent to there being no $F$-rational points. On the other hand, there exist curves with no $F$-rational points (hence incompressible) which still admit divisors defined over $F$ of degree $1$ - for instance, any curve over a finite field with no rational points will work (see here).
So let $X$ be an incompressible curve which admits a divisor $D$ of degree $1$, and consider the cycle $\alpha=X\times D\in\mathrm{CH}^1(X\times X)$. Its pushforward along $p_1$ is going to be equal to $\deg(D)X$ which is $1$ under your identifications, but since all the components of $\alpha$ are "horizontal", the pushforward along the other projection will vanish.