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.
EDIT
Of course there are similar examples involving chain complexes instead of spaces.
Note that in the case of the category $Ch(K)$ of chain complexes over a field $K$ the existence of a lifting is automatic. In fact, let $Ch'(K)$ be the full subcategory consisting of chain complexes whose boundary operators are zero: the composed functor $Ch'(K)\to Ch(K)\to Ho(Ch(K))$ is an equivalence of categories, so every diagram (of any shape) in the homotopy category can be lifted.
But of course this does not mean that we are silly to bother with chain complexes that have nonzero boundary operators. The point is that a commutative diagram in the homotopy category is not good for much. For example, Mayer-Vietoris sequences arise from certain square diagrams of chain complexes (say, those which are both pushouts and pullbacks), but two of these may be isomorphic as diagrams in the homotopy category and give different results.