How are you defining cuts? For the usual definition (which I give below), $-\alpha$ turns out to be $\{p:\exists r>0\, (-p-r\in \alpha)\}$, not $\{p:\exists r>0\, (-p-r\notin \alpha)\}$. I’ll go through this derivation in some detail, in hopes that you can adapt it to your situation.
To understand where the definition of $-\alpha$ comes from, you have to go back to the definition of the sum of two cuts. I’m going to assume the following definitions of cut and of addition of cuts:
$\alpha\subseteq \mathbb{Q}$ is a cut iff
$\;$
$\qquad(1)\qquad\varnothing\ne \alpha\ne\mathbb{Q}$;
$\qquad(2)\qquad\text{if }p; and
$\qquad(3)\qquad \alpha\text{ has no greatest element}$.
If $\alpha$ and $\beta$ are cuts, $\alpha+\beta\triangleq\{p\in\mathbb{Q}:\exists a\in \alpha\,\exists b\in \beta(p=a+b)\}$.
I’ll also assume that the embedding $^*$ of $\mathbb{Q}$ into the set of cuts has been defined so that $q^*=\{p\in\mathbb{Q}:p, so that $0^*=\{q\in\mathbb{Q}:q<0\}$ is intended to be the additive identity.
Clearly we want $-\alpha$ to be defined so that $\alpha+(-\alpha)=0^*$. Thus, I want $-\alpha$ to satisfy the following condition:
$\qquad\qquad\qquad\qquad$ for all $p\in\mathbb{Q}$, $p<0$ iff $\exists a\in \alpha\,\exists b\in -\alpha(p=a+b)$.
This is a bit hard to sort out, so let’s pretend for a moment that we’ve actually constructed $\mathbb{R}$ and look at a concrete example. Suppose that $\alpha$ is intended to be $\sqrt 2$: $\alpha=\{p\in\mathbb{Q}:p<\sqrt 2\}$. Clearly $-\alpha$ should be $-\sqrt 2$, or $\{p\in\mathbb{Q}:p<-\sqrt 2\}$. But $p<-\sqrt 2$ iff $-p>\sqrt 2$, so $-\alpha$ ought to be $\{p\in\mathbb{Q}:-p>\sqrt 2\}=\{p\in\mathbb{Q}:-p\notin\alpha\}$. This suggests that perhaps we should define $-\alpha$ in general to be $\{p\in\mathbb{Q}:-p\notin \alpha\}$. This almost works, but there’s a small problem if $\alpha$ is a rational cut $q^*$: in that case $\{p\in\mathbb{Q}:-p\notin q^*\}=\{p\in\mathbb{Q}:-p\ge q\}=\{p\in\mathbb{Q}:p\le -q\}$, which is not a cut, since it has a greatest element. We want $-(q^*)$ to be simply $(-q)^*$, i.e., $\{p\in\mathbb{Q}:p<-q\}$. There’s a simple (if slightly clumsy-looking) way around the problem: we simply set $-\alpha\triangleq\{p\in\mathbb{Q}:-p\notin\alpha\text{ AND }-p\text{ is not the least element of }\mathbb{Q}\setminus\alpha\}\;.\tag{1}$ It’s not hard to check that this really does define a cut. To check that it defines the right cut, suppose first that $a\in\alpha$ and $b\in-\alpha$. Then $-b\notin\alpha$, so $a<-b$, and therefore $a+b<0$, as desired.
Now suppose that $p$ is any negative rational; we need to show that there are $a\in\alpha$ and $b\in-\alpha$ such that $p=a+b$. This amounts to finding $a\in\alpha$ and $-b\in\mathbb{Q}\setminus\alpha$ such that $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$ and $a+b=p$ or, equivalently, $a-p=-b$.
Let $r=-p>0$; then we want to find $a\in\alpha$ and $-b\in\mathbb{Q}\setminus\alpha$ such that $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$ and $a+r=-b$. To give this some intuitive content, we’re trying to find rationals $a$ and $-b$ that are exactly $r$ apart and that ‘straddle’ the cut $\alpha$.
To do this, let $q\in\alpha$ and $s\in\mathbb{Q}\setminus\alpha$ be arbitrary. Since $\mathbb{Q}$ is non-Archimedean, there is an $n\in\mathbb{Z}^+$ such that $n>(s-q)/r$ and hence $q+nr>s$, so $\{n\in\mathbb{Z}^+:q+nr\notin\alpha\}\ne\varnothing$. Let $m=\min\{n\in\mathbb{Z}^+:q+nr\notin\alpha\}$; then $q+(m-1)r\in\alpha$ and $q+mr\notin\alpha$. Thus, if $q+mr$ is not the least element of $\mathbb{Q}\setminus\alpha$, we can simply take $a=q+(m-1)r$ and $-b=q+mr$ to get $a+r=-b$ with $a$ and $-b$ as desired.
If $q+mr$ is the least element of $\mathbb{Q}\setminus\alpha$, we have to be a little cleverer: in this case we let $a=q+\left(m-\frac12\right)r$ and $-b=q+\left(m+\frac12\right)r$. Clearly $a+r=-b\in\mathbb{Q}\setminus\alpha$, $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$, and $a, so $a\in\alpha$, again exactly as desired.
Notice that in the indented part of the argument we were actually showing that if $r$ is any positive rational, there are $a\in\alpha$ and $p\in-\alpha$ such that $-r=a+p$ or, equivalently, such that $a=-p-r$. In other words, we were showing that $\{p:\exists r>0\;(-p-r\in\alpha)\}\subseteq-\alpha$. On the other hand, if $p\in-\alpha$ by definition $(1)$, then $-p\notin\alpha$; let $a\in\alpha$ be arbitrary, set $r=-p-a$, and note that $r>0$ and $-p-r=a\in\alpha$, showing that $-\alpha\subseteq\{p:\exists r>0\;(-p-r\in\alpha)\}$. Thus, definition $(1)$ can be replaced by: $-\alpha\triangleq\{p\in\mathbb{Q}:\exists r>0\;(-p-r\in\alpha)\}\;.\tag{2}$