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 that 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.

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.)