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 ${\frak{g}}\subset \Lambda^2(\mathbb{R}^n)$ is the smallest subspace such that $R$ lies in $S^2({\frak{g}})$ (and this ${\frak{g}}$ is unique and easily computable from $R$), then ${\frak{g}}$ 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 ${\frak{g}}=\Lambda^2(\mathbb{R}^3)={\frak{so}}(3)$ but isn't the curvature of a symmetric space.

In addition, one needs 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 irreducible $n$-dimensional symmetric space with holonomy $G$.

The reason is that one then can define a Lie algebra structure on ${\frak{l}} = {\frak{g}}\oplus \mathbb{R}^n$ as follows: Let the bracket on ${\frak{g}}\subset{\frak{l}}$ be the usual bracket on ${\frak{g}}$; for $a\in{\frak{g}}\subset{\frak{so}}(n)$ and $x\in \mathbb{R}^n$, set $[a,x]=-[x,a]=ax$; and for $x$ and $y$ in $\mathbb{R}^n$, set $[x,y]= R(x{\wedge}y)$. (Here, we are regarding $R$ as a symmetric mapping $R:\Lambda^2(\mathbb{R}^n)\to \Lambda^2(\mathbb{R}^n)$, knowing that it has image in ${\frak{g}}\subset \Lambda^2(\mathbb{R}^n)$.) Then the assumption that $R$ lies in $S^2_B(\Lambda^2(\mathbb{R}^n))$ is just that $R(x{\wedge}y)z+R(y{\wedge}z)x+R(z{\wedge}x)y=0$ while the assumption that $R$ is invariant under $G$ implies that $a\ R(x{\wedge}y)-R(x{\wedge}y)\ a = R(ax{\wedge}y + x{\wedge}ay)$, and these are exactly the equations needed to verify that the bracket defined as above on ${\frak{l}}$ satisfies the Jacobi identity. Then the pair $({\frak{l}},{\frak{g}})$ is a symmetric pair, so that, by Cartan's construction, there is an $n$-dimensional Riemannian symmetric space $M=L/G$ with holonomy $G$ and having $R$ as its curvature operator.

*Added remark*: The OP wanted a criterion that didn't explicitly mention a Lie algebra $\frak{g}$, and one *can* do it this way: If one just defines $\frak{g}$ to be the image of $R$ in $\Lambda^2(\mathbb{R}^n)$ when one regards $R$ as a symmetric map from $\Lambda^2(\mathbb{R}^n)$ to itself, then the condition that
$$
R(w{\wedge}z)\ R(x{\wedge}y)-R(x{\wedge}y)\ R(w{\wedge}z)
= R\bigl(R(w{\wedge}z)x{\wedge}y\bigr) + R\bigl(x{\wedge}R(w{\wedge}z)y\bigr)
$$
hold for all $x,y,w,z\in\mathbb{R}^n$ implies both that $\frak{g}$ be closed under
Lie bracket (so that $\frak{g}$ is a Lie algebra *and* that $R$ be invariant under the action of the connected Lie subgroup $G\subset\mathrm{SO}(n)$ whose Lie algebra is $\frak{g}$. Thus, the above system of (quadratic) equations on $R$ is exactly the algebraic condition that $R\in S^2_B\bigl(\Lambda^2(\mathbb{R}^n)\bigr)$ be the curvature operator of a symmetric space. It's also, not surprisingly, the condition that the bracket defined above on $\frak{l}=\frak{g}\oplus\mathbb{R}^n$ satisfy the Jacobi identity. This may be more along the lines of what the OP had in mind with his question.

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 in several cases) that asserts that, in fact, such an $R$ *does* occur as the curvature operator at some point of a Riemannian metric with holonomy $G\subset\mathrm{SO}(n)$. I proved this using Cartan-Kähler theory for the exceptional cases $\mathrm{G}_2\subset\mathrm{SO}(7)$ and $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$ in *Metrics with exceptional holonomy*, Ann. of Math. (2) **126** (1987). It is obvious in the cases of $\mathrm{SO}(n)$ itself and $\mathrm{U}(n)\subset\mathrm{SO}(2n)$ and easy in the case of $\mathrm{SU}(n)\subset\mathrm{SO}(2n)$. I gave proofs (not in all detail, I admit) for the remaining cases $\mathrm{Sp}(n)$ and $\mathrm{Sp}(n)\mathrm{Sp}(1)$ in $\mathrm{SO}(4n)$ in *Classical, exceptional, and exotic holonomies: a status report*, Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), 93–165, Sémin. Congr., 1, Soc. Math. France, Paris, 1996, again using Cartan-Kähler theory.

However, it seems unlikely (though I don't know an explicit counterexample off the top of my head) that you could find a metric such that the curvature operator at *every* point is 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.)

*Added remark:* I have now checked, and, indeed, there is no $4$-dimensional Riemannian manifold with holonomy $\mathrm{SU}(2)$ such that the curvature operators $R_p$ at all points $p\in M$ are conjugate under $\mathrm{SO(4)}$. [Note that this condition is *a priori* weaker than the condition of being homogeneous; rather it's just the assumption that the germs of the metric at all points agree up to second order, i.e., that the metric is what is sometimes called 'curvature homogeneous'. (By comparison, there are many Riemannian $3$-manifolds such that all of the eigenvalues of the curvature are constant and so are 'curvature homogeneous', but that have no nontrivial Killing fields.)]