It's well-known (and straighforward to show) that $Cl_K[2]$ has order $2^{r-1}$ where $r$ is the number of prime factors of the discriminant. Since it's clear that there are infinitely many fundamental discriminants with precisely two prime factors, the answer is "yes".

Are there infinitely many $D$ such that $Cl_K[2] \cong \mathbb{Z} / 2\mathbb{Z}$?

It's well-known (and straighforward to show) that $Cl_K[2]$ has order $2^{r-1}$ where $r$ is the number of prime factors of the discriminant. Since it's clear that there are infinitely many fundamental discriminants with precisely two prime factors, the answer is "yes".

(EDIT: I added a quotation to clarify what exactly I am answering, since the original post hints at several distinct questions with rather different answers.)