Disease causing variants and Hardy-Weinberg Equilibrium
Hardy-Weinberg law makes a series of assumptions. One of them is the absence of selective effects. As you talk about disease, this assumption of neutrality is obviously not met.

At the moment of fecundation

Imagine for example that at the moment of the fecundation, the genotypes AA, Aa and aa are at Hardy-Weinberg equilibrium. Let x be the frequency of the allele A at the moment of fecundation, the frequencies of the genotypes AA, Aa and aa are $x^2$, $2x(1-x)$ and $(1-x)^2$, respectively.

After selection

As all genotypes aa die off, you're left with the A allele being at frequency $y=2x^2 + x(1-x)=x(1+x)$ and the frequencies of the genotypes AA, Aa and aa are $\frac{x^2}{x^2 + 2x(1-x)}=\frac{2x-2}{x-2}$, $\frac{2x(1-x)}{x^2 + 2x(1-x)}=\frac{x}{2-x}$, and $0$, respectively.

Expressing the genotype frequencies in terms of the new allele frequency

You can as well plug $y$ into those genotype frequencies to get the expression in terms of the genotype frequencies at birth ($y$; after selection). It gives you $\frac{1-\sqrt{4 y+1}}{\sqrt{4 y+1}-5}$, $\frac{2 \left(\sqrt{4 y+1}-3\right)}{\sqrt{4 y+1}-5}$ and $0$ for the genotypes AA, Aa and aa respectively. So, obviously, it is not at Hardy-Weinberg equilibrium.

Assumptions

You can make some more interesting calculations by leaving out the assumptions that all aa genotypes die at birth and you could use a Leslie matrix to describe the probability for an individual of each genotype to make to the next age. I also assumed that the a is completely recessive. You might want to release this assumption as well by computing the death of a fraction of the Aa genotypes.