There are two different types of creation and annihilation operators:
One is the bosonic set, which obeys the commutation relation $[a,a^\dagger]=1$ (and trivially $[a,a]=0=[a^\dagger,a^\dagger]$), and for which you can define quadratures via $q=\frac12(a+a^\dagger)$ and $p=\frac{1}{2i}(a-a^\dagger)$ which obey $[q,p]=i$, i.e. $a=q+ip$ and $a^\dagger=q-ip$ can be seen as the ladder operators for a harmonic oscillator with hamiltonian $H=\frac12(p^2+q^2)$.
The other is the fermionic set, which obeys the anticommutation relations $\{c,c^\dagger\}=1$, $\{c,c\}=0=\{c^\dagger,c^\dagger\}$, from which follows that $c^2=0=(c^\dagger)^2$.
The question incorrectly conflates properties of the two sets, i.e. the question as written asserts that 'the' creation operator obeys both $(a^\dagger)^2=0$ and $a^\dagger = q-ip$ where $q$ and $p$ are reasonably-behaved quadratures. There is no such operator: it's either fermionic, with the first property, or bosonic, with the second property.
However, OP's reference for that first property makes a much weaker claim: Gerry & Knight's eq. (2.45) does not require that $(a^\dagger)^2=0$ or that $a^2=0$; instead, it only requires that their expectation values over Fock states vanish, i.e.
$$
⟨n|a^2|n⟩=0=⟨n|(a^\dagger)^2|n⟩.
$$
These two properties are complex conjugates of each other, and they can be rigorously proved from the facts that
$$
a|n⟩=\sqrt{n}|n-1⟩
$$
so therefore
$$
a^2|n⟩=\sqrt{n(n-1)}|n-2⟩
$$
and thus
$$
⟨n|a^2|n⟩=\sqrt{n(n-1)}⟨n|n-2⟩=0.
$$
However, just because the expectation values of an operator vanish in the Fock basis (like they do for $a$ and $a^\dagger$ themselves) does not mean that the operator is identically zero, so e.g. $⟨0|a^2|2⟩ = \sqrt{2}\neq 0$ as a simple example.
