jmc

Because $p(x)$ has integer coefficients (in particular, because it has rational coefficients), the other root of $p(x)$ must be the radical conjugate of $3+\sqrt{7},$ which is $3-\sqrt{7}.$ Then, $p(x)$ must take the form $$p(x)=A(x-(3+\sqrt{7}))(x-(3-\sqrt{7}))$$for some nonzero constant $A$. This means that $$p(2)=A(-1+\sqrt{7})(-1-\sqrt{7})=-6A$$and $$p(3)=A(\sqrt{7})(-\sqrt{7})=-7A,$$so $$\frac{p(2)}{p(3)}=\frac{-6A}{-7A}=\boxed{\frac{6}{7}}.$$Alternatively, the roots are $3+\sqrt{7}$ and $3-\sqrt{7},$ so the sum of the roots is 6 and the product of the roots is $(3+\sqrt{7})(3-\sqrt{7})=9-7=2,$ so $$p(x)=A(x^2-6x+2)$$for some nonzero real number $A.$ Then $$\frac{p(2)}{p(3)}=\frac{A(-6)}{A(-7)}=\boxed{\frac{6}{7}}.$$