You should assume that each $A_n$ is a flat module over $R$.

Consider $A$ where each $A_n=R$ with basis element $x_n$, with multiplication $x_n x_m = (2-q)^{nm} x_{n+m}$. Since $nm = ((n+m)^2 - n^2 - m^2)/2$, $A$ is associative. The algebra $A' = A/(1-q)$ is simply $\mathbf{C}[x_1]$, so finitely generated, but $A$ is not finitely generated because $A/(2-q)$ has trivial multiplication.

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.

You should assume that each $A_n$ is a flat module over $R$.

Consider $A$ where each $A_n=R$ with basis element $x_n$, with multiplication $x_n x_m = (2-q)^{m} x_{n+m}$.

You should assume that each $A_n$ is a flat module over $R$.

Consider $A$ where each $A_n=R$ with basis element $x_n$, with multiplication $x_n x_m = (2-q)^{nm} x_{n+m}$. Since $nm = ((n+m)^2 - n^2 - m^2)/2$, $A$ is associative. The algebra $A' = A/(1-q)$ is simply $\mathbf{C}[x_1]$, so finitely generated, but $A$ is not finitely generated because $A/(2-q)$ has trivial multiplication.