I'll work in CW-complexes.
Consider the diagram S1+leftarrow∗+rightarrowS1+, where X+ denotes X with a disjoint basepoint added. Homotopy classes of maps ∗+toY are path components of Y, and homotopy classes of maps S1+toY are a choice of path component and a conjugacy class of element of pi1 of that path component. Therefore, if this diagram has a pushout in the homotopy category of based spaces it (co?)represents the functor sending Y to a choice of path component and a pair of conjugacy classes in the same path component.
Suppose we had a representing object X. Considering [X,S0], we find X has only two path components. So X=X0coprodX1 where X0 is the basepoint component. Each component can, up to homotopy equivalence, be constructed as a CW-complex with one zero-cell, some family of 1-cells, and some family of 2-cells.
Consider [X,K(pi,1)+] for pi a group. A cell description of X gives rise a description of this functor: an element of [X,K(pi,1)]+ is either trivial (if X1 maps to the basepoint) or is a conjugacy class of homorphism pi1(X1)topi, because the maps on X1 are not restricted to basepoint-preserving homotopies.
So it suffices to show that there are no groups G so that conjugacy classes of homomorphism Gtopi are in natural bijective correspondence with pairs of conjugacy classes of elements of pi.
EDIT: Fixed up the following argument.
Given such a group G, the identity map determines a pair of conjugacy classes [x],[y] in G, and choosing any representatives x and y determines a group homomorphism F2toG. Conversely, the conjugacy classes of the generators of F2 determine a map GtoF2 splitting this map up to conjugacy. This would imply that the natural tranformation sending simultaneous conjugacy classes of pairs to pairs of conjugacy classes is an inclusion, which is false, e.g. in the symmetric group on 3 letters there are 9 pairs of conjugacy classes and (assuming I counted correctly) 11 simultaneous conjugacy classes of pairs.
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