I don't know a reference, but I have some thoughts. I'm going to tackle this using just homology. I'll ignore gradings in the name of legibility.
There are three kinds of homology in this story: $H(M)$, $H(M,b M)$, and $H(b M)$. The last of these is self-dual; the other two are dual to each other. What I just said is literally true over $Q$, with 'dual to' meaning 'canonically isomorphic to the vector-space dual of'. Over $Z$ of course it's only true in a derived sense. One consequence of the derived statement is that, of the free abelian groups $H(M)/{torsion}$, $H(M,b M)/{torsion}$, and $H(b M)/{torsion}$, the last one is self-dual in the $Hom(-,Z)$ sense while the other two are dual to each other. Another is that the torsion part of $H(b M)$ is self-dual in the $Hom(-,Q/Z)$ sense and the torsion parts of the others are dual to each other in the same sense.
Then there are the other three players: the images of the three maps $H(M)\rightarrow H(M,b M)\rightarrow H(b M)\rightarrow H(M)$. Call them $A$, $B$, and $C$. (Each of the three can also be described as a kernel, or as a cokernel.) Here the rational story is very nice: $A$ is self-dual (more precisely the perfect pairing between $H(M)$ and $H(M, b M)$ when restricted to be a pairing between $H(M)$ and $A$ yields a pairing between $H(M)/ker=A$ and $A$ which is perfect). And $B$ and $C$ are dual to each other (more precisely, in the perfect self-pairing of $H(b M)$ the subspace $B$ is its own orthogonal complement, so that $B$ and $C=H(b M)/B$ are dual).
Over $Z$ things become murkier for $A$, $B$, and $C$. It's true that we get a pairing of $A$ with itself, and that this yields a nondegenerate pairing between the free abelian group $A/{torsion}$ and itself. But it's not perfect; it just injects $A$ into $Hom(A,Z)$. The same goes for pairing $B/{tors}$ with $C/{tors}$. And as for pairings of the torsion parts into $Q/Z$, the images just don't seem to behave well in general.
Update in response to Greg's update to the question: I think the following is basically what you're saying, but it can be said just algebraically and has nothing to do with choosing a splitting (i. e. identifying a quotient of $H(M)/tors$ with a subgroup). The sequence of free abelian groups $H(M)/tors\to H(M,bM)/tors\to H(bM)/tors$ is not exact. The image of $H(M)/tors\to H(M,bM)/tors$, which is the same as my $A/tors$, is not a summand in general; the kernel of $H(M,bM)/tors\to H(bM)/tors$ is a summand and consists of all elements such that some multiple is in $A/tors$; call this $A'$. So $A'$ contains $A/tors$ with finite index. And we get a perfect Z-pairing between $A/tors$ and $A'$. Likewise we get $B'$ and $C'$ containing $B/tors$ and $C/tors$ with finite index, and Z-dual to $C/tors$ and $B/tors$. I don't know what to say about all of this in relation to pairings into $Q/Z$.
Best Answer
The Hochschild-Serre spectral sequence