Understanding fail-stop subliminal channels
Answer to main question: Yes, it is indeed the way they work. Specifically, the rationale that in order to demonstrate whether or not a channel is subliminal-free, one needs to focus only on the instances in which A does not stop, is quite logical. Consider that, if A stops, then W realizes that a subliminal communication is taking place and the channel is not subliminal-free. But similarly, for the cases in which A does not stop, W still needs to decide if there is a subliminal channel. This is more clearly explained in "A Progress Report on Subliminal-Free Channels." (Check specifically Definition 3, although the prevoius sections help understanding the thought process of the authors.)

Answer to follow up question: TLDR; a fail-stop subliminal channel is  practical roughly when only one subliminal bit needs to be transmitted. Thus, it may make sense only in very restricted cases. The example by Desmedt is that in D day, it would have been enough to transmit a subliminal bit of "landing would not take place at Calais." When more bits need to be transmitted, it is very probably not practical, as its probability of going undetected goes down exponentially (and its capacity drops to $0$ just as quickly).

It can be naturally argued that the practicality of a subliminal channel is determined by its capacity, when the subliminal communication goes undetected.
The maximum capacity of a channel (in the absence of communication errors), is determined by its entropy. And given the previous answer, we know that a fail-stop subliminal channel is undetected as long as A does not halt, so we can use conditional entropy. 

The condition making A halt may be defined arbitrarily, but for the sake of simplicity, I stick to Desmedt's example (the following calculations are actually sketched in Desmedt's paper, although a bit succinctly...). In Desmedt's example, A stops if the last bit of the r value of the DSA signature matches $b \oplus c$, where $b$ is a secret (uniformly sampled) bit shared by A and B, and c is the subliminal bit to be transmitted. The event in which A does not stop implies that W sees a perfectly valid DSA signature that has been generated according to the protocol. Thus, he cannot differentiate it from a completely legitimate run. This is denoted with $\{w = 0\}$ (i.e., W ouputs "no subliminal information").

Lets generalize and assume that A wants to transmit u bits. I denote the bitstring to transmit subliminally with $c_i$, for every $i \in [0,2^u-1]$. 
Then, the capacity of one single execution of the protocol is:

$H(c_u|w = 0) = -\sum_{i\in[0,2^u-1]} p(c_i,w=0) \log \frac{p(c_i,w=0)}{p(w=0)} = -\sum_{i\in[0,2^u-1]} 2^{-2u} \log \frac{2^{-2u}}{2^{-u}}=u2^{-u}$

[Note that $p(c_i,w=0)$ is the event in which $c_i$ equals the $u$ least significant bits of $r$. Then, $p(c_i,w=0)=2^{-u} \cdot 2^{-u}$, as $c_i$ and $w=0$ are independent random variables, given that $b$ is chosen uniformly at random, and $r=(g^{k' \cdot k''} \bmod p) \bmod q$ is also drawn uniformly at random (because $k''$, chosen by W, is.)]

Now, the function $f(u)=u2^{-u}$ has a maximum at $1/\ln2$ and therefore the maximum capacity is $1/\ln2 \cdot 2^{-1/\ln2}=0.5307$.

Note that, if we want to analyze the case in which A tries to transmit t different signatures (restricting to the fail-stop strategy), each including $u$ bits of subliminal information, it is equivalent to having A try to transmit $tu$ bits in a single signature, which is again maximal when $tu = 1/\ln2$.