We can make example in genus 1 using any nontrivial extension of number fields $L/K$ with $[L:K]$ *odd*, and any non-CM elliptic curve $E$ over $K$. (By "non-CM" I mean no CM over $\overline{K}$.) More concretely, taking $K = \mathbf{Q}$, if $E$ is any non-CM elliptic curve over $\mathbf{Q}$ and $L$ is any number field of odd degree $> 1$ then there exist infinitely many a genus-1 curves $X$ over $L$ with Jacobian $E_L$ such that $X$ cannot descend to $\mathbf{Q}$ as an abstract curve.

To see this, suppose for such data $L/K$ and $E$ we knew that the natural restriction map
$$r:H^1(K,E) \rightarrow H^1(L,E)$$ is not surjective. A class in $H^1(L,E)$ corresponds to a genus-1 curve $X$ over $L$ such that its Jacobian is equipped with an identification with $E_L$. Suppose it happens that abstractly $X \simeq Y_L$ for a genus-1 curve $Y$ over $K$. Then $J(Y)$ is a $K$-descent of $J(X) = E_L$, which is to say that $J(Y)$ is a $K$-form of $E$ that becomes isomorphic to it over $L$. Since $E$ is not CM, so its geometric automorphic group is of order 2, the $K$-isomorphism class of $J(Y)$ corresponds to an element in $H^1(K,\mathbf{Z}/(2))$ that is split by the odd-degree $L/K$ and hence is trivial. That is, $J(Y)$ is $K$-isomorphic to $E$ and (with a moment's thought) $Y$ defines a class in $H^1(K,E)$ mapping to the class of $X$ in $H^1(L,E)$. Thus, if we could find a class in $H^1(L,E)$ not hit by $r$ then the corresponding $X$ would meet the desired conditions.

Pick an arbitrary nontrivial extension of number fields $L/K$ and any abelian variety $E \ne 0$ over $K$. I claim that the map $r$ as above is *never* surjective. To prove this, let $A$ be the abelian variety ${\rm{R}}_{L/K}(E_L)$ of dimension $[L:K]\dim(E) > \dim(E)$ given by Weil restriction, so there is a natural inclusion of $E$ as an abelian subvariety of $A$ with *non-zero* cokernel $B = A/E$. For a prime $\ell$ not dividing $[L:K]$, the map $r$ is on $\ell$-primary parts is injective with cokernel given by the $\ell$-primary part of $H^1(K,B)$ since the "norm" map $A \rightarrow E$ restricts to multiplication on $E$ by $[L:K]$ which is coprime to $\ell$.

Thus, it suffices to show that if $B$ is *any* nonzero abelian variety over $K$ (say with dimension $g > 0$) and if $\ell$ is any prime whatsoever then $H^1(K,B)[\ell]$ is infinite. (Such infinitude is no surprise since we impose no "local conditions" on this cohomology group such as unramifiedness away from a fixed finite set of places. The baby version is that $H^1(K,\mu_m) = K^{\times}/(K^{\times})^m$ is obviously infinite for any $m > 1$.) No doubt this infinitude can be proved using results known at the time of Tate's work on global Galois cohomology, but out of "laziness" let's instead argue using the refinement given by the product formula of Wiles.

Let $B^{\vee}$ be the dual abelian variety, and let $S$ be a finite set of non-archimedean places of $K$ consisting of *all* $\ell$-adic places and some places $v\nmid \ell$ where $B^{\vee}[\ell]$ is locally split (so $B^{\vee}[\ell](K_v)$ has order $\ell^{2g}$ for such $v$).
Let $\delta$ be the $\mathbf{F}_{\ell}$-dimension of $B(K)/(\ell)$. Let $S^{(\ell)}$ denote the set of places in $S$ away from $\ell$. Note that by Chebotarev applied to the splitting field of $B^{\vee}[\ell]$ over $K$ we can make $S^{(\ell)}$ as large as we wish. A straightforward application of the Kummer sequence and the Wiles product formula gives that the image in $H^1(K,B)[\ell]$ of the subgroup of $H^1(K,B[\ell])$ satisfying the Selmer condition away from $S$ has $\mathbf{F}_{\ell}$-dimension at least
$$g[K:\mathbf{Q}]-\delta+2g|S^{(\ell)}|.$$
Now by making $S^{(\ell)}$ as large as we wish, we can make this lower bound arbitrarily large. Thus, $H^1(K,B)[\ell]$ is infinite.

respectingthe given $K$-structure on the abelian surface. In principle, it might happen that the p.p. abelian surface over $L$ admitsanother$K$-descent as a p.p. abelian surface (i.e., with a $K$-structure different from the given one on the abelian surface), so the associated curve over $L$ would then be defined over $K$. So to make an example in this way one needs a stronger "does not descend to $K$" property. Perhaps you were already aware of this, in which case all I'm doing is clarifying your comment. – grp Oct 5 '12 at 0:06