>[...] suggests to me that what I wrote is indeed sufficient due to the separability.

This is not true. Indeed, by (say) Theorem 2.3.1 in Bogachev's book, the operator $R$ must be nuclear. 

>If this is false, then I am happy with answers saying it is true but you need these extra requirements on $\mathcal{B}$ or $R$.

As noted in Remark 3.11.24, with further references there, "the class of all nuclear, symmetric and
nonnegative operators between $X$' and $X$ coincides with the class of the covariance
operators of Gaussian measures on $X$ precisely when $X$ is a space of type $2$."