This is really a response to Karl's beautiful example; I'm posting it as an "answer" only because there isn't enough room to leave it as a comment.
The condition on conductor ideals is one that I had come across by thinking about the dual picture.
Namely, let $f:Y\rightarrow X$ be a finite map of 1-dimensional proper and reduced schemes over an algebraically closed field $k$. Then $Y$ and $X$ are Cohen-Macaulay by Serre's criterion, so the machinery of Grothendieck duality applies. In particular, the sheaves
$f_*O_Y$ and $f_*\omega_Y$ are dual via the duality functor $\mathcal{H}om(\cdot,\omega_X)$, as are
$O_X$ and $\omega_X$. Here, $\omega_X$ and $\omega_Y$ are the ralative dualizing sheaves of
$X$ and $Y$, respectively. Thus, the existence of a trace morphism $f_*O_Y\rightarrow O_X$
is equivalent by duality to the existence of a pullback map on dualizing sheaves $\omega_X\rightarrow f_*\omega_Y$.
In the reduced case which we are in, one has Rosenlicht's explicit description of the dualizing sheaf: for any open $V$ in $X$, the $O_X(V)$-module $\omega_X(V)$ is exactly the set of
meromorphic differentials $\eta$ on the normalization $\pi:X'\rightarrow X$ with the property that $$\sum_{x'\in \pi^{-1}(x)} res_{x'}(s\eta)=0$$
for all $x\in V(k)$ and all $s\in O_{X,x}$.
It is not difficult to prove that if $C$ is the conductor ideal of $X'\rightarrow X$
(which is a coherent ideal sheaf on $X'$ supported at preimages of non-smooth points in $X$),
then one has inclusions
$$\pi_*\Omega^1_{X'} \subseteq \omega_X \subseteq \pi_*\Omega^1_{X'}(C).$$
Since $X'$ and $Y'$ are smooth, so one has a pullback map on $\Omega^1$'s, our question
about a pullback map on dualizing sheaves boils down the following concrete question:
When does the pullback map on meromorphic differentials $\Omega^1_{k(X')}\rightarrow \pi_*\Omega^1_{k(Y')}$ carry the subsheaf $\omega_X$ into $\pi_*\omega_Y$?
By looking at the above inclusions, I was led to conjecture the necessity of conductor ideal containment as in my original post. As Karl's example shows, this containment is not sufficient.
Here is Karl's example re-worked on the dual side:
Set $B:=k[x,y]/(xy)$ and $A:=k[u,v]/(uv)$ and let $f:A\rightarrow B$ be the $k$-algebra map
taking $u$ to $x^2$ and $v$ to $y$. Writing $B'$ and $A'$ for the normalizations, we have
$B'$ and $A'$ as in Karl's example, and the conductor ideals are $(x,y)$ and $(u,v)$.
Now the pullback map on meromorphic differentials on $A'$ is just
$$(f(u)du,g(v)dv)\mapsto (2xf(x^2)dx,g(y)dy).$$
The condition of being a section of $\omega_A$ is exactly
$$res_0(f(u)du)+res_0(g(v)dv)=0,$$
and similarly for being a section of $\omega_B$. But now we notice that
$$res_0(2xf(x^2)dx)+res_0(g(y)dy) = 2 res_0(f(u)du) + res_0(g(v)dv) = res_0(f(u)du)$$
if $(f(u)du,g(v)dv)$ is a section of $\omega_A$. Thus, as soon as $f$ is not
holomorphic (i.e. has nonzero residue) the pullback of the section
$(f(u)du,g(v)dv)$, as a meromorphic differential on $B'$, will NOT lie in the subsheaf $\omega_B$.
Clearly what goes wrong is that the ramification indices of the map $f:A'\rightarrow B'$
over the two preimages of the nonsmooth point are NOT equal. With this in mind, I propose the following addendum to my original number 4):
In the notation of 4) above and of Karl's post, assume that $f'(C_A)=C_B^e$
for some positive integer $e$. Then the trace map $B'\rightarrow A'$ carries $B$
into $A$.
Certainly this rules out Karl's example. I think another way of stating the condition is that the map $f':Spec(B')\rightarrow Spec(A')$ should be "equi-ramified" over the nonsmooth locus of $Spec(A)$, i.e. that the ramification indices of $f'$ over all $x'\in Spec(A')$ which map to the same nonsmooth point in $Spec(A)$ are all equal.
Is this the right condition?
The answer to your first question is obviously no, as you point out: the homomorphism $A\rightarrow A/I$ is essentially never flat. It is OK under your hypotheses ($f$ injective, $A$ noetherian, $B$ finitely generated) if you assume moreover that $B$ is a domain. A reference (in french, sorry) is Bourbaki, Algèbre Commutative X, §4, no. 3, Corollaire.
The hypothesis that $B$ is a domain is necessary for stupid reasons: just take $B=A\times A/I$ with $A/I$ Cohen-Macaulay, again this is not flat over $A$.
Best Answer
The "Theorem" isn't true with both rings just normal, or just CM, or even normal and CM. Let $A = k[[x,y,z]]/(xz-y^2) \cong k[[a^2,ab,b^2]]$ and let $B = k[[a,b]]$, with $f$ the natural inclusion. The dimensions add up as they must, since $f$ is module-finite. In this case finite flat dimension is the same as finite projective dimension, but $B$ does not have finite projective dimension over $A$.
I don't expect that any addition of assumptions $(R_i)$ would help.