Here is a recipe for constructing some examples.
Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots. Suppose that $f$ has the following two properties:
The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity.
We have $-\overline{f(-\overline x)} = f(x)$.
(For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.)
Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$. The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points). As a result, the only possible such automorphisms are the identity and $y \mapsto -y$.
Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$.
Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$. This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$.
However, this curve cannot be defined over $\Bbb R$. The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$. In terms of group cohomology, this is determined by an extension $$ 1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1. $$ Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear.
In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$. To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$. This automorphism is the hyperelliptic involution, and so the extension is nontrivial.
I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.