Given that $$\mathbb{R}P^{\infty} = B O(1) = K(\widehat{O(1)}, 1)$$ $$\mathbb{C} P^{\infty} = B U(1) = K( \widehat{U(1)}, 2)$$ is there any way to make sense of $$\mathbb{H}P^{\infty} = B Sp(1)$$ in a similar manner using the representation theory of the nonabelian group $Sp(1) \cong Spin(3) \cong SU(2)$?

There is a principal bundle However, this implies that it 'is' an EilenbergMac Lane space after rationalization, $$\mathbb{H}P^\infty\simeq_{\mathbb{Q}}K(\mathbb{Z},4).$$ 


Any group $G$ has a classifying space $BG$. It can be a finite group, an infinite discrete group a Lie group or any topological group. The construction is always the same, find a contractible space, usually called $EG$, with a free continuous action by $G$. Then $BG$ is the quotient $EG/G$. One then has a fibration (in fact a principal bundle) $$G\to EG \to BG.$$ The long exact sequence in homotopy groups gives an isomorphism of $\pi_i G$ and $\pi_{i+1}BG$ (homotopy groups of $EG$ are all $0$). When $G$ is discrete, this gives $\pi_1BG=G$ and $BG$ is aspherical (no higher homotopy groups). If $G$ is $S^1$, then, by definition, $BS^1=CP^{\infty}$. Moreover $\pi_1 S^1 = \pi_2 CP^{\infty}=\mathbb Z$ are the only nontrivial groups of $S^1$ and $CP^{\infty}$. So $CP^{\infty} = K(Z,2)$. The $3$sphere is a Lie group, a.k.a SU(2), but has lots of nontrivial homotopy groups. So $BS^3$ is not $K(\mathbb Z,4)$ even though $\pi_4 BS^3 = \pi_3 S^3 = \mathbb Z$. For $n\ge2$, $K(A,n)$ is defined only when $A$ is a abelian group. 


I claim that, in the equivalences you stated, duality has nothing to do with it. Specifically, if your viewpoint were correct, then for finite $G$, $BG(1)$ would be noncanonically isomorphic to $K(G,1)$. In fact, I claim that they are canonically isomorphic. Furthermore, the $\mathbb Z$ in $\mathbb CP^{\infty}$ is canonically isomporphic to the fundamental group, not the character group, of $U(1)$. Argument: Since both these spaces represent functors, it suffices to consider the underlying functors. Eilenberg MacLane spaces correspond to cohomology functors. It is easy to prove using Cech cohomology that cohomology with coefficients in $G$ naturally classifies principal $G$bundles. This, of course, is exactly what the classifying space classifies  not dual to what the classifying space classifies. $\mathbb CP^{\infty}$: There is an exact sequence $0\to\mathbb Z \to \mathbb C^+ \to \mathbb C^\times\to 0$, giving a map $H^1(X,\mathbb C^\times)\to H^2(X,\mathbb Z)$. The image is discrete while the kernel, a quotient of $H^1(X,\mathbb C^{+})$, is connected, so the map is exactly the quotient by the connected component of the identity. $H^1(X,\mathbb C^\times)$ classifies principal $\mathbb C^\times$ spaces. Continuously moving the bundle around in it corresponds to continuously deforming the bundle. These bundles up to derivation are exactly what $BU(1)$ classifies. EDIT: Idea/sketch for a general proof of this equivalence: Let $G_n$ be the group of principal $G$bundles on $S^{n}$. Then for some reason this should be equivalent to $\pi_{n1}(G)$. Now, the values everywhere of good functors on the category of CW complexes (specifically, representable ones) depend only on the values they take on spheres. So suppose a group had only one nontrivial $G_n$. The principal $G$spaces functor would then be equivalent to $H^n(X,G_n)$, giving an equivalence of classifying spaces. 


This is not really an answer to the question posed but seems to be of relevance to people interested in the question (and is directly related to the case $BSU(2)\cong_{\mathbb{Q}}K(\mathbb{Z},4)$ mentioned by Mark Grant). There is a sequence of groups for which the classifying spaces are rationally products of EilenbergMaclane spaces: namely $BU(n)$. The $i$th Chern class is an element of $H^{2i}(X,\mathbb{Z})$ and hence can be thought of as homotopy class of map to a $K(\mathbb{Z},2i)$ space. Therefore you get a map $$c_1\times\cdots\times c_n\colon BU(n)\to \prod_{i=1}^nK(\mathbb{Z},2i)$$ which turns out to be a rational homotopy equivalence. I learned this trick from Atiyah & Bott (http://www.jstor.org/stable/10.2307/37156), Section 2. I guess the same should work for $SU(n)$ when you leave out the $c_1$ factor. 

