You don't even need $sigma^2$ to construct such a variable.
Let $Z$ be $+1$ with probability $q$, and $-1$ with probability $1-q$, and independent of $X$.
Let $Y = e^XZ$. This squashes X to the positive reals preserving order, and then may change the sign.
$y_1$ and $y_2$ have the same ordering as $x_1$ and $x_2$ when the greater value isn't negated. That is, if $x_1 gt x_2$, then $sign((x_1 - x_2)(y_1-y_2)) = sign(y_1)$. If $x_1lt x_2$, then $sign((x_1 - x_2)(y_1-y_2)) = sign(y_2)$. So, $E[sign((x_1 - x_2)(y_1-y_2))] = E[Z].$
Choose $Z$ to have average value $p$ (set $q=frac{(p+1)}2$), and then $X$ and $Y$ have Spearman Rank Correlation Coefficient $p$.
Actually, there is a little ambiguity (to me) about whether you allow $x_1 = x_2$, which I ignored above. If under your definition, the rank correlation of $X$ with itself is $alpha$, then the rank correlation of $X$ with $Y$ is $alpha p$, and you can get any value in $[-alpha,alpha]$.
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