I think I have worked out my confusion so I thought I should post it as an answer. The original question was not well-posed; hopefully this will help anyone else who has similar misunderstandings.
Keeping spin in the picture, the space of states for an individual quark is the tensor product of the three-dimensional flavour space with the two-dimensional spin representation of SU(2). However, this is not considered as a representation of SU(3) x SU(2), rather as a representation of SU(6) containing this product as a subgroup. In other words you can rotate flavours into spins and vice versa. As a representation of SU(6), the tensor cube of this standard 6-dimensional representation decomposes into pieces, one of which is the symmetric cube. This is an irreducible 56-dimensional representation. This decomposes under the subgroup SU(3) x SU(2) into a direct sum of two pieces: one is the SU(3)-decuplet tensored with the (4-dimensional) spin-3/2 representation of SU(2), one is the adjoint representation of SU(3) (the octet) tensored with the spin-1/2 representation of SU(2) (and indeed 10 x 4 + 8 x 2 = 56).
When you forget about spin, the spin-1/2 octet that appears here is actually a mixture of terms from the two SU(3)-octets in the decomposition of the tensor cube from the original question. In other words, the actual wavefunction of a proton is a sum of two terms, one involving terms from one SU(3)-octet and one involving terms from the other (both tensored with suitable spin wavefunctions).
I found these notes of Jiří Chýla very helpful in sorting out my misunderstanding: