Here is a solution to the hint:
First of all, note that since all the ideals in question contain $\mathfrak a$, we may replace
$A$ by $A/\mathfrak a$, and so assume that $\mathfrak a = 0$; this simplifies the notation somewhat.
Next, the condition that $\mathfrak b_1 \cap \cdots \cap \mathfrak b_r = 0$ is equivalent
to the requirement that the natural map $A \to A/\mathfrak b_1 \times \cdot \times
A/\mathfrak b_r$ (the product ot the natural quotient maps) is injective, while the
condition that $\mathfrak b_i$ is irreducible is equivalent to the statement that
if $I$ and $J$ are non-zero ideals in $A/\mathfrak b_i$, then $I \cap J \neq 0$ also.
Now suppose given our two irreducible decompositions of $0$. Choose $i$ as in the hint,
and set $I_j := \mathfrak b_1 \cap \cdots \cap \mathfrak b_{i-1} \cap \mathfrak c_j\cap \mathfrak b_{i+1}
\cap \cdots \cap \mathfrak b_r,$ for each $j =1,\ldots,s$.
Then $I_1\cap \cdots \cap I_s = 0$ (since it is contained in the intersection
of the $\mathfrak c_j$, which already vanishes).
Now we recall that $A$ embeds into the product of the $A/\mathfrak b_{i'}$.
Note that $I_j$ is contained in $\mathfrak b_{i'}$ for $i'\neq i$. Thus,
if we let $J_j$ denote the image of $I_j$ in $A/\mathfrak b_i$, then
we see that the image of $I_j$ under the embedding
$A \hookrightarrow A/\mathfrak b_1\times\cdots\times A/\mathfrak b_i \times \cdots \times A/\mathfrak b_r$
is equal to $0 \times \cdots \times J_j \times\cdots \times 0$.
Thus the intersection of the images of the $I_j$, which is the image
of the intersection of the $I_j$ (since we looking at images under an embedding)
is equal to $0\times \cdots \times (\bigcap J_j) \times \cdots \times 0.$
Thus, since the intersection of the $I_j$ is equal to $0$, we see
that $\bigcap J_j = 0.$ But $\mathfrak b_i$ is irreducible, and so
one of the $J_j = 0$. Equivalently, the corresponding $I_j = 0.$
This proves the hint.
(I think the exercise should be a fairly easy deduction from the hint. The
statement that $r = s$ at least follows directly, using the hint together with
minimality of the two decompositions.)
The definition of $f:R\to S$ being faithfully flat that I first saw is that $S\otimes_R-$ is exact and faithful (meaning that $S\otimes_R M=0$ implies $M=0$). I'm not sure exactly what your definition of "faithfully flat" is, but it looks like you're happy with "flat and surjective on spectra." You get flatness for free from étaleness, so I'll show that the extra faithfulness condition implies surjectivity on spectra.
Upon tensoring with $S$, $f$ becomes $f\otimes_R id_S:S\cong S\otimes_R R\to S\otimes_R S$, given by $s\mapsto s\otimes 1$. This is injective since it is a section of multiplication $S\otimes_R S\to S$. By flatness of $S$, this shows that $S\otimes_R \ker f=0$, so $\ker f=0$. So I'll identify $R$ with a subring of $S$.
Let $\mathfrak p\subseteq R$ be a prime ideal. We wish to show that there is a prime $\mathfrak q\subseteq S$ such that $\mathfrak q \cap R=\mathfrak p$. Let $K$ be the kernel of the morphism $R/\mathfrak p\to S/\mathfrak p S$ of $R$-modules. Upon tensoring with $S$, this morphism becomes injective (as before, it's a section of the multiplication map $S/\mathfrak p S\otimes_R S/\mathfrak p S\to S/\mathfrak p S$), so by flatness of $S$, we have $S\otimes_R K=0$, so $K=0$. This shows that $\mathfrak p S \cap R=\mathfrak p$ (if the intersection were any larger, $K$ would be non-zero). So $\mathfrak p$ generates a proper ideal in the localization $(R\setminus \mathfrak p)^{-1}S$. Let $\mathfrak q\subseteq (R\setminus \mathfrak p)^{-1}S$ be a maximal ideal containing $\mathfrak p$. This corresponds to some prime ideal $\mathfrak q$ (slight abuse of notation to use the same letter) of $S$ which contains $\mathfrak p$ but does not intersect $R\setminus \mathfrak p$, so $\mathfrak q\cap R=\mathfrak p$.
See also exercise 16 of Chapter 3 of Atiyah-Macdonald.
Best Answer
EDIT: Don't bother reading my partial solution. Brian Conrad pointed out that an easier way to do what I did is to use the equivalent definition of formally unramified in terms of Kähler differentials. And later on, fpqc posted below a complete solution passed on by Mel Hochster, who got it from Luc Illusie, who got it from ???.
OLD ANSWER: Here is a half-answer. I'll prove half the conclusion, but on the plus side I'll use only half the hypothesis! Namely, I will prove that if $R \to S_{\mathfrak{p}}$ is formally unramified for all primes $\mathfrak{p} \subset S$, then $R \to S$ is formally unramified.
Let $A$ be an $R$-algebra, and let $I \subseteq A$ be a nilpotent ideal. Given $R$-algebra homomorphisms $f,g \colon S \to A$ that become equal when composed with $A \to A/I$, we must prove that $f=g$. Fix $\mathfrak{p}\subset S$. Then the localizations $A_{\mathfrak{p}} := S_{\mathfrak{p}} \otimes_{S,f} A$ and $S_{\mathfrak{p}} \otimes_{S,g} A$ of $A$ (defined viewing $A$ as an $S$-algebra in the two different ways) are naturally isomorphic, since adjoining the inverse of an $a \in A$ to $A$ automatically makes $a+\epsilon$ invertible for any nilpotent $\epsilon$ (use the geometric series). Now $f$ and $g$ induce $R$-algebra homomorphisms $f_{\mathfrak{p}},g_{\mathfrak{p}} \colon S_{\mathfrak{p}} \to A_{\mathfrak{p}}$ that become equal when we compose with $A_{\mathfrak{p}} \to A_{\mathfrak{p}}/I A_{\mathfrak{p}}$. Since $R \to S_{\mathfrak{p}}$ is formally unramified, this means that $f_{\mathfrak{p}} = g_{\mathfrak{p}}$. In other words, for every $s \in S$, the difference $f(s)-g(s)$ maps to zero in $A_{\mathfrak{p}}$ for every $\mathfrak{p}$. An element in an $S$-module that becomes $0$ after localizing at each prime ideal of $S$ is $0$, so $f(s)=g(s)$ for all $s$. So $f=g$.