The issue seems to be about the definition of coherence. The OP states that multimode light could be perfectly coherent. In the context of temporal coherence the term multimode means multiple frequencies. However, as explained by the OP, a light source with multiple frequencies can produce a reduced fringe visibility.
Well, the definition of first order coherence is based on fringe visibility. Therefore, multiple frequencies do not produce a perfectly coherent optical field. In fact, the coherence length of an optical source is inversely proportional to width of the frequency spectrum of the source. If the spectrum is a Dirac delta function, the coherence length would be infinite, which means the source is perfectly coherent. However, physical light sources always have a spectrum with a finite width. Therefore, the coherence length of such sources are finite. It means that multimode light in this context is not perfectly coherent.
For a full treatment of optical (first order) coherence, you can consult the book by Mandel and Wolf "Optical coherence and quantum optics." Emil Wolf is known for having developed a comprehensive theory of optical coherence.
Based on the comments, it may be useful to say a bit more about coherent states and whether one can say that a coherent state is always perfectly coherent. Firstly, a coherent state is nominally an eigenstate of the annihilation operator. If we incorporate the additional degrees of freedom, we have
$$ \hat{a}_s(\mathbf{k}) |\alpha\rangle = |\alpha\rangle \alpha_s(\mathbf{k}) , $$
where the eigenvalue function $\alpha_s(\mathbf{k})$ is a complex valued function with a spin index $s$ and wave vector dependence $\mathbf{k}$ taken over from the annihilation operator. Each eigenvalue function is associated with a unique eigenstate, which is the coherent state. One can turn the annihilation operator into the annihilating part of a field operator by a Fourier transform and then the eigenvalue function will become a function of space and time. Hence
$$ \hat{E}(\mathbf{x},t) |\alpha\rangle = |\alpha\rangle \mathbf{E}(\mathbf{x},t) , $$
where the eigenvalue function $\mathbf{E}(\mathbf{x},t)$ become an electric field that parameterized the coherent state. For the sake of this discussion, let's focus only on time. So we consider
$$ \hat{E}(t) |\alpha\rangle = |\alpha\rangle \mathbf{E}(t) . $$
If we now ask whether the coherent state is coherent according to the definition of first order coherence, as determined by the observation of fringe visibility, then the obvious way to answer this question is to perform an experiment. Such an experiment would involve an interferometer, such as a Mach-Zender, consisting of two beam splitters that respectively separate the state into two paths and then recombines them again. Prior to the second beam splitter, a relative phase is introduced between the two paths. Without going through the detailed calculations, I can hopefully convince you what the interferometer does to the coherent state is to convert its parameter function into a superposition of the parameter function and a shifted version of that parameter function. In other words,
$$ \mathbf{E}(t) \rightarrow \frac{1}{\sqrt{2}} \left[ \mathbf{E}(t)+\mathbf{E}(t+\tau) \right] . $$
With a slight abuse of notation, we may represent the coherent state as
$$ |\alpha_{out}\rangle = \left| \mathbf{E}(t)+\mathbf{E}(t+\tau) \right\rangle , $$
where I'm discarding the normalization to simplify the expression.
Next we need to measure the intensity of the output to see the fringes. For this purpose, we use a number operator, which would give us the expectation value for the number of photons, which is proportional to the intensity. Hence
$$ \langle \hat{n} \rangle = \langle \alpha_{out}|\hat{n} |\alpha_{out}\rangle = |\mathbf{E}(t)+\mathbf{E}(t+\tau)|^2 . $$
So, we see that the final result is precisely the same expression that we would expect to find in classical optics, from which it then follows that the visibility of the fringes would depend on the coherence properties of the electric field, which is also the parameter function of the coherent state. In general, it would not be perfectly coherent, as for instance in the case of an electric field with a finite bandwidth. I am not sure what Loudon is referring to, but it is clear that a coherent state does not necessarily represent a state that is perfectly coherent according to first order coherence theory.
