I think I just figured this out. For cloning you want to do the following:
$(A|p\rangle+B|q\rangle)|0\rangle$ goes to $(A|p\rangle+B|q\rangle)(A|p\rangle+B|q\rangle)$ where $|p\rangle$ and $|q\rangle$ are orthogonal.
For stimulated emission one gets:
$(A|p\rangle+B|q\rangle)|0\rangle = A|p\rangle|0\rangle+B|q\rangle|0\rangle$ goes to $A|p\rangle|p\rangle+B|q\rangle|q\rangle$ which is different from the cloning case.
Hence, it does not violate the no-cloning theorem.
It also seems to me (judging from Mandel and Wolf) that $|p\rangle$ and $|q\rangle$ must be in the momentum basis for stimulated emission. One cannot use an arbitrary basis.
Update:
Not withstanding the recognition that @StevenSagona has given my old answer, for which I am grateful, I am not convinced that it is the best way to understand the issue. It has been several years since I provided this answer and in the meantime I have developed a different understanding. So, let me present that understanding.
The question is how stimulated emission works. One often finds the idea that the medium must have some intelligence that can determine the incoming state and then reproduces it. Assuming the input state is some Fock state where all photons have the angular spectrum $F$, then such a process would be represented by:
$$ |n_F\rangle \rightarrow |(n+1)_F\rangle . $$
It assumes that the stimulated process can be modeled by an individual creation operator $\hat{a}_F$, which is not a unitary operator. Instead, the process should be represented by a unitary operator. Is it possible to find such a unitary operator $\hat{U}$ for this process? If such a unitary operator does exist, it would imply
$$ \langle n_G|n_F\rangle = \langle G,F\rangle^n
= \langle n_G|\hat{U}^{\dagger}\hat{U}|n_F\rangle
= \langle (n+1)_G|(n+1)_F\rangle = \langle G,F\rangle^{(n+1)} . $$
It means that, either $|\langle G,F\rangle|=1$ or $|\langle G,F\rangle|=0$, which is inconsistent with the requirement that $F$ and $G$ be arbitrary angular spectra. Here, we have reproduced the no-cloning theorem, but for a slightly different scenario. So, we see that the first expression does not give a suitable representation of the stimulated emission process.
Then how does stimulated emission work? The thing is that tendency for the emission to have the same state as the incoming state is a result of a bosonic enhancement. It is not that the medium only reproduces the incoming states. It produces everything that the structure allows, but the part the corresponds to the incoming state receives a bosonic enhancement.
To model this process, let's assume that a spontaneous emission would produce a single photon state:
$$ |\psi\rangle = |1_a\rangle C_a + |1_b\rangle C_b + |1_c\rangle C_c + ... , $$
where the subscripts represent the different modes$^{\star}$ that can be radiated and the $C$'s are arbitrary coefficients. To compare it with the incoming state $|n_F\rangle$, we can recast the expression as
$$ |\psi\rangle = |1_F\rangle \alpha + |1_{\bar{F}}\rangle \beta , $$
where $\bar{F}$ is the orthogonal complement of $F$, and $\alpha$ and $\beta$ are the associated coefficients so that $|\alpha|^2+|\beta|^2=1$. In the stimulated process one would nominally get
$$ |\psi\rangle|n_F\rangle = |1_F\rangle|n_F\rangle \alpha
+ |1_{\bar{F}}\rangle|n_F\rangle \beta . $$
However, this state is not properly normalized. If we look at the two terms, we see that $|1_{\bar{F}}\rangle|n_F\rangle$ is normalized,
$$ \langle 1_{\bar{F}}|1_{\bar{F}}\rangle \langle n_F|n_F\rangle = 1 $$
but $|1_F\rangle|n_F\rangle$ is not a normalized state. Instead it becomes
$$ |1_F\rangle|n_F\rangle = |(n+1)_F\rangle\sqrt{n+1} . $$
The factor $\sqrt{n+1}$ represents the bosonic enhancement. It causes the latter term to dominate over the term with the orthogonal complement. So the stimulated emission process is better represented as
$$ |n_F\rangle \rightarrow |(n+1)_F\rangle\sqrt{n+1} + |1_{\bar{F}}\rangle|n_F\rangle , $$
where I assumed $\alpha=\beta=1$ and ignored the overall normalization. Obviously this is not a simple cloning of the incoming state.
$\star$ Here the term mode is used in its true sense as a solution of the equations of motion satisfying the boundary conditions.
