Let $R=\mathbb C[q^{\pm 1}]$ and let $A$ be a graded (possibly non-commutative) $R$-algebra, $A=\oplus\_{n=0}^\infty A\_n,$ where $A\_0=R$ and all $A\_n$'s are free $R$-modules. 
Then $A'=A/(q-1)$ is a graded algebra over $R/(q-1)=\mathbb C$ and so 
$A$ can be thought as a deformation of $A'.$ I couldn't find a definition of a flat deformation of algebras but I would imagine this being an example of it, correct?
(Assume that $A'$ is commutative if it helps.) 


Main question: Is it possible that $A'$ is finitely generated while $A$ is not (as algebras over $\mathbb C$ and $R$ respectively)?



Finite generation of flat deformations of algebras
To say that a deformation is flat, you should assume that $A$ is flat over $R$, e.g. each $A\_n$ is a free $R$-module of finite rank.


The answer to the main question is no, even if $A$ is commutative. Consider $A$ where each $A\_n=R$ with basis element $x\_n$, with multiplication defined by the formula $$x\_n x\_m = (2-q)^{nm} x\_{n+m}.$$ Since $2nm = (n+m)^2 - n^2 - m^2$, the algebra $A$ is associative. The algebra $A' = A/(1-q)$ is simply $\mathbf{C}[x\_1]$, so is finitely generated. However, $A$ is not finitely generated because $A/(2-q)$ has trivial multiplication and therefore e.g. the ideal generated by $x\_n$ ($n>1$) is not finitely generated. 


In deformation theory, one usually considers limits of deformations over Artinian algebras. In your setting, it could be more natural to consider complete algebras over the power series ring $\mathbf{C}[[q-1]]$. In this setting, the answer to the main question is probably positive.


