Thanks for your hints so far. The question was to some extend already discussed here: (https://physics.stackexchange.com/questions/109343/eigenstate-of-field-operator-in-qft). <PERSON>, I give a more verbose and down-to-earth answer to my question. There is also a remark why I was confused first...
Eigenstates of the annihilation field as classical states
We assume a real valued scalar quantum field $A(x)$ of the form (neglecting normalization)
\begin{equation} \hat A(x) = \int \hat a^{*}(k)e^{ikx} \, d^{3} k + \int \hat a(k)e^{-ikx} \, d^{3} k = \hat A_c(x) + \hat A_a(x) \end{equation}
which is split in a creation and annihilation part. 669
We are looking for a eigenstate $| \psi \rangle$ of the annihilation operator $\hat A_a(x)$. Such a state is the tensor product of single mode coherent states: $| \psi \rangle = \otimes_k |\alpha(k) \rangle_k$, since every single mode state in the tensor product is an eigenstate of the corresponding annihilation operator $a(k)$ with eigenvalue $\alpha(k)$. Hence $\hat A_a(x) | \psi \rangle = \int d^{3} k' e^{-ik'x} \hat a(k')\otimes_k |\alpha(k) \rangle_k =\int d^{3} k' e^{-ik'x} \alpha(k') \otimes_k|\alpha(k) \rangle_k = \psi(x) | \psi \rangle$
with the eigenvalue $\psi(x) = \int d^{3} k e^{-ikx} \alpha(k)$. The eigenstate $| \psi \rangle$ is also called coherent and can be written as \begin{equation} | \psi \rangle = \otimes_k e^{\alpha(k) \hat a^{}(k)} |0 \rangle = e^{\int d^{3}k \alpha(k) \hat a^{}(k) } |0 \rangle = e^{\int d^{3}x \hat A_c(x) \psi(x)} |0 \rangle = e^{\int d^{3}x \hat A(x) \psi(x)} |0 \rangle \end{equation} The first part of the above equation is more or less by definition true (refer to single mode coherent states) the second part is valid since all $\hat a^{*}(k)$ commute, the third part is valid since (Fourier expansion of field and eigenvalue)
\begin{equation} \int d^{3}x \hat A_c(x) \psi(x) = \int \int d^{3} k d^{3} q \, \int e^{i(k - q)x} d^{3}x \, \alpha(q) \hat a^{} (k) = \int d^{3} k \alpha(k) \hat a^{}(k) \end{equation}
(The space integral results in a delta function $\delta(k-q)$). The last part of the defining equation above is true since $\hat A_a |0 \rangle = 0$ and hence $e^{\int d^{3}x \hat A_a(x) \psi(x)} |0 \rangle = 0$.
Corollary: $\langle \psi |\hat A_a(x) |\psi \rangle = \langle \psi |\psi(x)|\psi \rangle = \psi(x) \langle \psi |\psi \rangle = \psi(x)$. 66