How can the integral of a bra and ket in equation (6) of Berry's 1984 paper be a vector?

[Dickfore]

It seems the notion I was lacking was that nabla acting on a ket vector in Hilbert space yields another vector in hilbert space (I hope).

Yes it does. Think about the partial derivative:
$$\frac{\partial}{\partial R_i} \, \vert n(\mathbf{R}) \rangle \equiv \lim_{t \rightarrow 0} \frac{\vert n(R_1, \ldots, R_i + t, \ldots, R_n) \rangle - \vert n(R_1, \ldots, R_i, \ldots, R_n) \rangle}{t}$$
Since the expression in the limit is always a linear combination of two kets, it must be a ket (that is what is meant by the Hilbert space being a linear space). Thus, every partial derivative is a ket. You may treat these partial derivatives as COVARIANT components of a gradient in some n dimensional metric space. This metric space may be curved, depending on the nature of the parameters.

 

[Jazzdude]

It seems the notion I was lacking was that nabla acting on a ket vector in Hilbert space yields another vector in hilbert space (I hope).

No, it yields a vector of hilbert space vectors. You have to understand that there are two vector spaces here, one for states, one for parameters.