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One thing to be aware of is that, even when liftings to strict diagrams exist, their non-uniqueness is a serious matter. For example, consider the square pushout diagram in which $S^n$ maps to a point, and to $D^{n+1}$ by inclusion; the fourth space is $S^{n+1}$. Compare this with a different diagram in which the four spaces are the same and three of the maps are the same but the map $S^n\to D^{n+1}$ is replaced by a constant map (to some point in the boundary). This is again strictly commutative, and it yields an isomorphic diagram in $Ho(Top)$, but the one square is "highly connected" while the other is not: the first square gives a map from the homotopy fiber of $S^n\to \star$ (which is $S^n$) to the homotopy fiber of $D^{n+1}\to S^{n+1}$ (which is homotopy equivalent to $\Omega S^{n+1}$) that induces an isomorphism of $H_n$, while the second square induces a map between precisely the same two homotopy fibers which, being a constant map, cannot possibly induce an isomorphism.