Given a convex quadrilateral $ABCD$ inscribed in a circle with $\angle A = 60^\circ$, $BC = CD = 1$, and the intersections of rays $AB$ and $DC$ at point $E$, and rays $BC$ and $AD$ at point $F$, we aim to find the perimeter of quadrilateral $ABCD$ given that the perimeters of triangles $BCE$ and $CDF$ are integers.

First, we note that $\angle BCD = \angle BAC = 60^\circ$ since $ABCD$ is cyclic and $\angle A = 60^\circ$. Let $a$ and $b$ be the angles at $E$ and $F$ respectively such that $a + b = 120^\circ$.

We consider the triangle $CDF$. Since $\angle CDF = 60^\circ$, we can use the properties of a 30-60-90 triangle to find that the perimeter of $\triangle CDF$ is an integer. Similarly, the perimeter of $\triangle BCE$ is also an integer.

Using the Law of Sines in $\triangle CDF$, we have:
$$\frac{1}{\sin(b - 30^\circ)} = \frac{y}{\sin 60^\circ} \implies y = \frac{\sin 60^\circ}{\sin(b - 30^\circ)}$$
$$\frac{1}{\sin(b - 30^\circ)} = \frac{x}{\sin(150^\circ - b)} \implies x = \frac{\sin(150^\circ - b)}{\sin(b - 30^\circ)}$$

Summing these, we get:
$$x + y = \frac{\sin 60^\circ + \sin(150^\circ - b)}{\sin(b - 30^\circ)} = \frac{\sqrt{3}/2 + \cos b + \sqrt{3}\sin b}{\sqrt{3}\sin b - \cos b} = 3$$

Solving for $\cos b$ and $\sin b$, we find:
$$\cos b = \frac{3\sqrt{3}}{14}, \quad \sin b = \frac{13}{14}$$

Using the Law of Sines in $\triangle ABD$, we have:
$$\frac{AD}{\sin a} = \frac{BD}{\sin 60^\circ} = 2 \implies AD = 2\sin a$$
$$AB = 2\sin b = \frac{13}{7}$$

Since $a + b = 120^\circ$, we have:
$$\sin a = \sin(120^\circ - b) = \frac{\sqrt{3}\cos b + \sin b}{2} = \frac{9}{28} + \frac{13}{28} = \frac{11}{14}$$
$$AD = 2\sin a = \frac{11}{7}$$

Thus, the perimeter of quadrilateral $ABCD$ is:
$$AB + BC + CD + DA = \frac{13}{7} + 1 + 1 + \frac{11}{7} = \frac{38}{7}$$

The answer is $\boxed{\frac{38}{7}}$.