Why there is no unique "recipe" for quantization of a classical theory?

1. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $\mathbb{R}^{6N}$ phase space dynamics to be recoverable from the thermodynamic limit $N\to\infty$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one.

2. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $f(x,p)\mapsto \hat{f}$. The Groenewold-van Hove theorem says that there is no such map such that

   1. $f\mapsto \hat{f}$ is linear.
   2. $[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$ holds for all observables $f,g$.
   3. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible.
   4. $p(\hat{f}) = \hat{p(f)}$ for all polynomials $p$,

   meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $\hbar^2$, geometric quantization instead restricts the allowed inputs $f$ to the quantization map and drops the fourth property.

   Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $f\mapsto \hat{f}$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question.

Firstly, it should be stressed that different quantisation approaches to a classical theory will lend different insights. Secondly, one quantisation method for a system may be particularly advantageous over others depending on what one would like made manifest.

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There is a prototypical example of this. Consider for example the action of a classic string,

$$S = -\frac{1}{4\pi\alpha'}\int d^2 \sigma \, \sqrt{-h} \, h^{ab}\partial_a X_\mu \partial_b X^\mu.$$

Even among canonical quantisation, there are different gauges one can choose which will offer different insights. The light-cone gauge allows one to arrive at the spectrum of the string the fastest but covariance of the theory is manifest with the conformal gauge. The light-cone gauge is able to eliminate the diffeomorphism and Weyl redundancies.

Now, a second approach to the classical string is BRST quantisation. One can classify states as being BRST-exact or BRST-closed in the same sense of being closed or exact for differential forms and thus introduce BRST cohomology analogous to de Rham cohomology.

The physical Hilbert space is identified with this BRST cohomology, and it is a theorem (proved in 4.4 of Polchinski) that,

$$\mathscr{H}_{\mathrm{CQ}} = \mathscr{H}_{\mathrm{BRST}} = \mathscr{H}_{\mathrm{light-cone}}$$

which is to say the Hilbert space matches up with the one obtained from canonical quantisation as well as light-cone quantisation. Thus, although the BRST method has some boons, it offers an equivalent description of the system.

As for proving equivalences in more general cases, I hope another member of the SE can offer insights.

The answers above are great, but they don't address the last question of yours, so here goes.

$-$ Can spin networks be used to quantize QCD?

$-$ Only if it is coupled to gravity.

The spin network basis is uncountable. The inner product space is thus nonseparable and is unable to describe a well-defined quantum mechanical system.

The beautiful reason why this works for gravity is because the kernel of the diffeomorphism constraint (appropriately quantized as an operator on the spin network space) of GR is actually a separable Hilbert space $\mathcal{K}$, which is usually called the kinematical Hilbert space of LQG. In other words, because LQG is background independent, the "excessive size" of the spin-networks inner product space is just gauge, the true Hilbert space being separable.

This will also work for gravity + $SU(3)$ Yang-Mills system (QCD). But it won't work for QCD in the flat Minkowski background. Background independence really makes the difference here.