Here's an attempt, assuming the target category is finite dimensional vector spaces and the diagram is finite acyclic.
1) Fill in the diagram so it's triangulated (that is, add the compositions of all arrows). Add kernels and cokernels to every arrow in the diagram. Add the natural arrows between these new objects (e.g. the kernel of $f$ maps into the kernel of $g\circ f$). Iterate this process until it terminates (which it does, as there are finitely many arrows, and thus only finitely many subquotients of objects they induce).
2) In this new diagram, tabulate all exact paths. (This step seems to me to have high time and space complexity, unfortunately.) By exact path, I mean an infinite long exact sequence, for which all but finitely many objects are zero.
3) Given an exact path $$0\to A_1\to \cdots \to A_i\to \cdots \to A_n\to 0$$ write the equation $$\sum_{i=1}^n (-1)^i \dim~ A_i=0.$$
4) Solve the resulting linear system for the $\dim~ A_i$'s.
Since we've added kernels and cokernels, solving the system tells us about injectivity and surjectivity of all the arrows. Furthermore, as far as I can tell any (non-negative) solution to this linear system is realizable as a diagram, so this algorithm gives us any information we could get by diagram chasing.
Greg has given a good summary of this method in his answer below--he takes me to task about my assertion that any non-negative integer solution to this linear system is realizable, so I will sketch an argument (again requiring the diagram to be finite acyclic). We also assume the diagram is "complete" in the sense that step 1) has been completed.
Partially order the objects in the diagram by saying $A\leq B$ iff $A$ is a subquotient of $B$; inductively, this is the same as saying that $B\leq B$ and $A\leq B$ if there exists $C\leq B$ such that $A$ injects into $C$ or $C$ surjects onto $A$ in this diagram. We identify isomorphic objects; this is a poset by acyclicity. We proceed by induction on the number of objects in the diagram. The one-object diagram is easy, so we do the induction step.
By finiteness the diagram contains a longest chain; choose such a chain and consider its minimal element $X$. All maps into or out of $X$ are surjections or injections, respectively, so split $X$ off of every object that maps onto it or that it maps into as a direct summand. The diagram with $X$ removed gives the induction step.
A well-written discussion of the group completion can be found on pp. 89--95 of
J.F. Adam: Infinite loop spaces, Ann. of Math. studies 90 (even though he only
discusses a particular group completion of a monoid). In particular you
assumption of commutativity comes in under the assumption that $\pi_0(M)$ is
commutative which makes localisation with respect to it well-behaved
(commutativity is not the most general condition what is needed is some kind of
Øre condition).
In any case if you really want conclusions on the homotopy equivalence level I
think you need to put yourself in some nice situation for instance requiring
that all spaces be homotopy equivalent to CW-spaces. If you don't want that you
should replace homotopy equivalences by weak equivalences, if not you will
probably find yourself in a lot of trouble. In any case I will assume that we
are dealing with spaces homotopy equivalent to CW-complexes.
Starting with 1) a first note is that your conditions does not have to involve
an arbitrary ring $R$. It is enough to have $R=\mathbb Z$ and one should
interpret the localisation in the way (for instance) Adams does:
$H_\ast(X,\mathbb Z)=\bigoplus_\alpha H_\ast(X_\alpha,\mathbb Z)$, where
$\alpha$ runs over $\pi_0(X)$, and a $\beta$ maps $H_\ast(X_\alpha,\mathbb Z)$
to $H_\ast(X_{\alpha\beta},\mathbb Z)$. Then your group completion condition is
that the natural map $\mathbb Z[\pi_0(Y)]\bigotimes_{\mathbb Z[\pi_0(X)]}
H_\ast(X,\mathbb Z)\rightarrow H_\ast(Y,\mathbb Z)$ should be an isomorphism.
This then implies the same for any coefficient group (and when the coefficient
group is a ring $R$ you get your condition). (Note that for this formula to even make sense we need at least associativity for the action of $\pi_0(X)$ on the homology. This is implied by the associativity of the Pontryagin product of $H_*(X,\mathbb Z)$ which in turn is implied by the homotopy associativity of the H-space structure.)
Turning now to 1) it follows from standard obstruction theory. In fact maps into
simple (hope I got this terminology right!) homotopy types, i.e., spaces for
which the action of the fundamental groups on the homotopy groups is trivial (in
particular the fundamental group itself is commutative). The reason is that the
Postnikov tower of such a space consists of principal fibrations and the lifting
problem for maps into principal fibrations is controlled by cohomology groups
with ordinary coefficients. Hence no local systems are needed (they would be if
non-simple spaces were involved). The point now is that H-spaces are simple so
we get a homotopy equivalence between any two group completions and as
everything behaves well with respect to products these equivalences are H-maps.
Addendum:
As for 2) it seems to me that this question for homotopy limits can only be solved under supplementary conditions. The reason is that under some conditions we have the Bousfield-Kan spectral sequence (see Bousfield, Kan: Homotopy limits, completions and localizations, SLN 304) which shows that $\varprojlim^s(\pi_s X_i)$ for all $s$ will in general contribute to $\pi_0$ of the homotopy limit. As the higher homotopy groups can change rather drastically on group completion it seems difficult to say anything in general (the restriction to cosimplicial spaces which the OP makes in comments doesn't help as all homotopy limits can be given as homotopy limits over $\Delta$. Incidentally, for homotopy colimits you should be in better
shape. There is however an initial problem (which also exists in the homotopy limit case): If you do not assume that the
particular group completions you choose have any functorial properties it is not
clear that a diagram over a category will give you a diagram when you group
complete. This can be solved by either assuming that in your particular
situation you have enough functoriality to get that (which seems to be the case
for for instance May's setup) or accepting "homotopy everything" commutative
diagrams which you should get by the obstruction theory above. If this problem
is somehow solved you should be able to conclude by the Bousfield-Kan spectral
sequence $\injlim^\ast H_*(X_i,\mathbb Z)\implies
H_*(\mathrm{hocolim}X_i,\mathbb Z)$. We have that localisation is exact and
commutes with the higher derived colimits so that we get upon localisation a
spectral sequence that maps to the Bousfield-Kan spectral sequence for $\{Y_i\}$
and is an isomorphism on the $E_2$-term and hence is so also at the convergent.
As for 3) I don't altogether understand it. Possibly the following gives some
kind of answer. For the H-space $\coprod_n\mathrm{B}\Sigma_n$ which is the
disjoing union of classifying spaces of the symmetric groups its group
completion has homotopy groups equal to the stable homotopy groups of spheres
which shows that quite dramatic things can happen to the homotopy groups upon
group completion (all homotopy groups from degree $2$ on of the original space
are trivial).
Best Answer
I'm not sure if this will still be helpful, but here is my understanding of the Quillen model. Everything correct that I write below, I learned from John Francis. (Probably in the same lecture that Theo mentioned in his comment above.) Any mistakes are not his fault---more likely an error in my understanding.
Before we begin: Quillen v Sullivan.
As others have mentioned, Quillen gets you a DG Lie algebra, where as the Sullivan model will get you a commutative DG algebra. As you write, the passage from one to the other is (almost) Koszul duality. Really, a Lie algebra will get you a co-commutative coalgebra by Koszul duality, and a commutative algebra will get you a coLie algebra. You can bridge the world of coalgebras and algebras when you have some finiteness conditions--for instance, if the rational homotopy groups are finite-dimensional in each degree. Then you can simply take linear duals to get from coalgebras to algebras.
A way to find Lie algebras.
So where do (DG) Lie algebras come from? There is a natural place that one finds Lie algebras, before knowing about the Quillen model: Lie algebras arise as the tangent space (at the identity) of a Lie group $G$.
Now, if you're an algebraist, you might claim another origin of Lie algebras: If you have any kind of Hopf algebra, you can look at the primitives of the Hopf algebra. These always form a Lie algebra.
(Recall that a Hopf algebra has a coproduct $\Delta: H \to H \otimes H$, and a primitive of $H$ is defined to be an element $x$ such that $\Delta(x) = 1 \otimes x + x \otimes 1.$)
One link between the algebraist's fountain of Lie algebras, and the geometer's, is that many Hopf algebras arise as functions on finite groups. If you are well-versed in algebra, one natural place to find Lie algebras, then, would be to take a finite group, take functions on that group, then take primitives.
A cooler link arises when a geometer looks at distributions near the identity of $G$ (which are dual to 'functions on $G$') rather than functions themselves. This isn't so obviously the right thing to look at in the finite groups example, but if you believe that functions on a Lie group $G$ are like de Rham forms on $G$, then you'd believe that something like 'the duals to functions on $G$' (which are closer to vector fields) would somehow safeguard the Lie algebra structure. The point being, you should expect to find Lie structures to arise from things that look like 'duals to functions on a group'. So one should take 'distributions' to be the Hopf algebra in question, and look at its primitives to find the Lie algebra of 'vector fields.'
A (fantastical) summary of the Quillen model.
Let us assume for a moment that your space $X$ happens to equal $BG$ for some Lie group, and you want to make a Lie algebra out of it. Then, by the above, what you could do is take $\Omega X = \Omega B G = G$, then look at the primitives of the Hopf algebra known as `distributions on $\Omega X$'.
Now, instead of considering just Lie groups, let's believe in a fantasy world (later made reality) in which all the heuristics I outlined for a Lie group $G$ will also work for a based loop space $\Omega Y$. A loop space is `like a group' because it has a space of multiplications, all invertible (up to homotopy). Moreover, any space $X$ is the $B$ (classifying space) of a loop space--namely, $X \cong B \Omega X$. So this will give us a way to associate a Lie algebra to any space, if you believe in the fantasy.
Blindly following the analogy, `functions on $\Omega X$' is like cochains on $\Omega X$, and the dual to this (i.e., distributions) is now chains on $\Omega X$. That is, $C_\bullet \Omega X$ should have the structure of what looks like a Hopf algebra. And its primitives should be the Lie algebra you're looking for.
What Quillen Does.
So if that's the story, what else is there? Of course, there is the fantasy, which I have to explain. Loop spaces are most definitely not Lie groups. Their products have $A_\infty$ structure, and correspondigly, we should be talking about things like homotopy Hopf algebras, not Hopf algebras on the nose. What Quillen does is not to take care of all the coherence issues, but to change the models of the objects he's working with.
For instance, one can get an actual simplicial group out of a space $X$ by Kan's construction $G$. This is a model for the loop space $\Omega X$, and this is what Quillen looks at instead of looking only at $\Omega X$, which is too flimsy. From this, taking group algebras over $\mathbb{Q}$ and completing (these are the simplicial chains, i.e., distributions), he obtains completed simplicial Hopf algebras. Again, instead of trying to make my fantasy precise in a world where one has to deal with higher algebraic structures (homotopy up to homotopy, et cetera) he uses this nice simplicial model. To complete the story, he takes level-wise primitives, obtaining DG Lie algebras.
Edit: This is from Tom's comment below. To recover a $k$-connected group or a $k$-connected Lie algebra from the associated $k$-connected complete Hopf algebra, you need $k \geq 0$. And $k$-connected groups correspond to $k+1$-connected spaces. This is why you need simply connected spaces in the equivalence.
I'm not sure I gave any 'high concept' as to 'why Quillen's construction works', but this is at least a road map I can remember.