One must distinguish between quantum/classical on the string world-sheet and in spacetime.
Both of your statements are basically correct, but should read something like "CFT theory is the space of classical solutions to the spacetime equations of string theory" and "Quantization of the
the world-sheet sigma model of a string theory gives rise to a CFT."
In a little more detail,
the sigma-model describing string theory propagation on some manifold M is a 2-dimensional
quantum field theory which in order to describe a consistent string theory must be a conformal
field theory. The "classical limit" of this 2-dimensional field theory is a limit in which some
measure of the curvature of M is small in units of the string tension. To construct a CFT one
must solve the sigma-model exactly, including world-sheet quantum effects.
The coupling constants of the sigma-model are fields in spacetime such as the metric $g_{\mu \nu}(X(\sigma))$ on $M$ where $X: \Sigma \rightarrow M$ define the embedding of the string world-sheet $\Sigma$ into $M$. Now there
is also a spacetime theory of these fields. You can think of it as a ``string field theory". At low-energies it can sometimes be usefully approximated by a theory of gravity coupled to some finite number
of quantum fields, but in full generality it is a theory of an infinite number of quantum fields. Roughly speaking, each operator in the CFT gives rise to a field in spacetime. The spacetime string field theory lives
in 10 dimensions for the superstring or 26 dimensions for the bosonic string and it also has a classical limit. The classical limit is $g_s \rightarrow 0$ where $g_s$ is
a dimensionless coupling constant. It appears in perturbative string theory as a factor which
weights the contribution of a Riemann surface by the Euler number of the surface. It can also be
thought of as the constant (in spacetime) mode of a scalar spacetime field known as the dilaton.
The main point is that there are two notions of classical/quantum in string theory, one involving
the world-sheet theory, the other the spacetime theory. In order to avoid confusion one must be clear which is being discussed. Unfortunately string theorists often assume it is clear from the context.
In response to the further question about the space of string fields, I would suggest that you have a look at the introductory material in http://arXiv.org/pdf/hep-th/9305026. You may also find http://arXiv.org/pdf/hep-th/0509129 useful. I should add that while string field theory has had some success recently in the description of D-brane states, it is not widely thought to be a completely satisfactory definition of non-perturbative string theory.
To the best of my knowledge of the literature on this topic, the answer is: not really (a few exceptions appear below). Let me first give a rigorous statement of the problem: the physical equation $\beta = 0$ can be expressed as
$\sum_{k=0}^{\infty} \epsilon^k \beta_k = 0$,
where $\epsilon$ is a physical parameter one imagines is "small," and where $\beta_k$
is a symmetric two tensor depending on a Riemannian metric which arises as a certain universal expression in the curvature tensor and its derivatives, and which moreover obeys a natural scaling law and a natural "homogeneity" in terms of the number of derivatives of $g$ which are required to express it. These terms $\beta_k$ are computable in theory, although my understanding is that this has only been done on a piecemeal basis for the first few terms. There are many references one can easily find which do this for this and other sigma models. The first two terms are $\beta_0 = \mbox{Vol}(M) g$, $\beta_1 = \mbox{Rc}$, the term $\beta_2$ is a quadratic expression in the curvature tensor. Considering the equation up to order $k$ in the $\epsilon$ power expansion is referred to in physics literature as "up to $k$-loops."
So, with this background, there are at least two interesting questions one can ask:
1) Given $N \in \mathbb N \cup \{\infty\}$, can one construct on a given manifold a solution to $\sum_{k=0}^N \epsilon^k \beta_k = 0$?
2) Can one construct on a given manifold a one-parameter family of Riemannian metrics satisfying $\frac{\partial g}{\partial t} + \sum_{k=0}^N \epsilon^k \beta_k = 0$?
The original question was related to 1) above, but 2) is as important in the physics literature (AFAIK). Of course, the case $N=1$ of question 1) corresponds to solving the usual Riemannian Einstein equation (although one may need to justify ignoring the term $k=0$ in some cases, which physicists have ways of doing). In the Kähler setting the metric can come from the Calabi-Yau theorem as the poster stated. For question 2), the case $N=1$ corresponds to the (properly normalized) Ricci flow equation.
Now that I have properly stated the question, let me just say that I focus on studying elliptic/parabolic equations on manifolds, and for a while got interested in exactly this question. After much literature digging, I found only a few rigorous mathematical results:
http://arxiv.org/abs/0904.1241
http://arxiv.org/abs/1108.0526
http://arxiv.org/abs/1205.6507
The first paper considers the case $N=2$ of question 2). In this case the operator is still second-order, and so in special settings the equation can be rendered parabolic, and then known techniques can be applied. This paper is certainly interesting, although the techniques cannot really be extended to $N > 2$. The last two papers address the homogeneous setting.
Moving beyond $N = 2$ is, IMHO, extremely difficult from a PDE point of view, since one has very little information on the form of the terms $\beta_k$ beyond the general qualitative statements I made above. For instance, an interesting side question is: do there exist arbitrarily large values of $k$ for which $\beta_k$ corresponds to an elliptic operator of a Riemannian metric? This is true for $k=1$, but I believe it is very unlikely to be true for higher $k$ since I believe it is "known" in the physics sense that all of the terms arising in $\beta_k$ for $k > 1$ are at least quadratic combinations of curvature and derivatives. I.e., one cannot expect a term like $\Delta \mbox{Rc}$ to show up, which would be elliptic.
One can imagine considering the case $N > 2$ on homogeneous spaces, where in principle the equations reduce to a system of ODE, but again one does not have much in the way of understanding the form of the general term $\beta_k$.
It is also not inconceivable that if one starts with say a Ricci-flat metric, and is allowed to choose $\epsilon$ very small with respect to this metric, that one can perturb it to a solution to the higher order equations, but again without at least a little more hard data on the form of the $k$-th term this seems tricky.
EDIT: In response to Robert's comments, yes, I meant $\mbox{Vol}(M) g$ (fixed above). Also, let me give a reference:
http://www.sciencedirect.com/science/article/pii/0550321389904227
This paper computes the terms $\beta_3$ and $\beta_4$ (The formula for $\beta_4$ covers half a page of terms, and each of these is shorthand for a complicated curvature expression already. Written out fully would take 2 pages at least!). As far as whether one can expect the $\beta$ functions to be simpler on a Kähler manifold, I can't speak to the physics well enough to really say. From a mathematical point of view, I don't really see any advantage beyond the fact that of course each term could be expressed in terms of a Kähler potential.
The case of symmetric spaces is probably more tractable. Looking for solutions within the class $\nabla \mbox{Rm} \equiv 0$ could certainly simplify the terms $\beta_k$ greatly. My understanding of how these terms are derived is extremely fuzzy, but let me speak anyways: I believe the different summands in $\beta_k$ correspond in some sense to different sized "loop diagrams" (see the above reference for some examples), and moreover the qualitative behavior of such terms (i.e. number of derivatives of $\mbox{Rm}$ which appear is (somehow) related to the topology of the diagram. One main difficulty in performing these calculations to derive the terms $\beta_k$ is the sheer number of diagrams which must be considered, which grows wildly with $k$. If, on the other hand, one could a priori rule out a large number of such diagrams (by assuming $\nabla \mbox{Rm} \equiv 0$), perhaps these calculations become more tractable, or at least some stronger qualitative statements could be made.
Lastly, the quantity $\epsilon$ (which, also note is usually $\alpha'$ in physics literature) is, I believe, meant to be small but fixed. Physically it apparently is meant to represent the "string tension."
Best Answer
A physicist would answer this question as follows. (Everything I'll say can be expressed in a way that the purest of mathematicians would understand, but that translation would take a lot of work, so I'll only do it on demand.)
In physics we have units of mass ($M$), length ($L$) and time ($T$).
In special relativity we have a fundamental constant $c$, the speed of light, with units $L/T$. Thus, in special relativity, any length determines a time and vice versa.
In quantum mechanics we have a fundamental constant $\hbar$, Planck's constant, with units $ML^2/T$. $\hbar / c$ has units $M L$. Thus, in theories involving both special relativity and quantum mechanics, any mass determines an inverse length, and vice versa.
Relativistic quantum field theory involves both special relativity and quantum mechanics. Thus, in relativistic quantum field theory, if we have a particle of some mass $m$, it determines a length, namely $\hbar / c m$. This is called the Compton wavelength of that particle.
Some physicists may prefer to use $h = 2 \pi \hbar$ in the definition of the Compton wavelength, but this isn't important here: what really matters is that when you have a relativistic quantum field theory with a particle of some given mass, it will have a preferred length scale and will thus not be invariant under scale transformations, hence not under conformal transformations... unless that mass is zero, in which case the Compton wavelength is undefined.
What is the meaning of the Compton wavelength? Here's a rough explanation. In quantum mechanics, to measure the position of a particle, you can shoot light (or some other kind of wave) at it. To measure the position accurately, you need to use light with a short wavelength, and thus a lot of energy. If you try to measure the position of a particle with mass $m$ more accurately than its Compton wavelength, you need to use an energy that exceeds $mc^2$. This means that the collision is energetic enough to create another particle of the kind whose position you're trying to measure. So, in relativistic quantum field theory, we should imagine any particle as part of a 'cloud' of 'virtual' particles which can become 'real' if you try to measure its position too accurately. The size of this cloud is not sharply defined, but it's roughly the Compton wavelength. So, the theory of this particle is not invariant under conformal transformations.