I suppose that $S^2_B(\Lambda^2(\mathbb{R}^n))$ means the subspace of $S^2(\Lambda^2(\mathbb{R}^n))$ that satisfies the Bianchi identity, i.e., the kernel of the natural map $$ S^2(\Lambda^2(\mathbb{R}^n))\longrightarrow \Lambda^4(\mathbb{R}^n). $$
Certainly, you'd need that, if $L\subset \Lambda^2(\mathbb{R}^n)$ is the smallest subspace such that $R$ lies in $S^2(L)$ (and this $L$ is unique and clearly computable from $R$), then $L$ has to be the Lie algebra of a closed, connected subgroup $G\subset \mathrm{SO}(n)$.
This is not enough, though, as the case $n=3$ shows. (The generic $R$ in this case has $L=\Lambda^2(\mathbb{R}^3)={\frak{so}}(3)$ but isn't the curvature of a symmetric space.) You also need that $R$ be invariant under the action of this $G$ on $S^2(\Lambda^2(\mathbb{R}^n))$. Conversely, if this invariance holds, then $R$ is the curvature of an $n$-dimensional symmetric space with holonomy $G$.
Your other question about characterizing curvature operators of Riemannian manifolds with reduced holonomy is not as easy to answer. You can, of course, tell when $R$ lies in $S^2({\frak{g}})\subset S^2(\Lambda^2(\mathbb{R}^n))$ for a given Lie algebra ${\frak{g}}\subset{\frak{so}}(n)=\Lambda^2(\mathbb{R}^n)$, but then knowing whether there is a metric with holonomy $G$ whose curvature tensor takes the value $R$ at some point is a little tricky.
In the special case that $\frak{g}$ acts irreducibly on $\mathbb{R}^n$ and can be the holonomy of a Riemannian metric that is not locally symmetric, there is a result (I guess, due to me) that asserts that, in fact, such an $R$ does occur as the curvature operator at some point of a Riemannian metric with holonomy $G\subset\mathrm{SO}(n)$. However, it is unlikely (though I don't know an explicit counterexample off the top of my head) that you could find such a metric such that the curvature operator at every point was equivalent to $R$ up to $G$-conjugacy. (Probably, this is not possible even for $\mathrm{SU}(2)\subset\mathrm{SO}(4)$, but I'd have to think about it to be sure.)