= Solution
\b[The literal assertion needs $q\ne p$.] If $q=p$, the two <geodesic circles> coincide, have equal tangent lines at every point, and do not intersect transversally; their intersection is not finite. We prove that <close geodesic circles intersect twice> for nearby distinct centres, and in fact find exactly two intersections.
Choose a strongly convex normal neighbourhood about $p$ and a sufficiently small $r>0$ so that every circle considered and all <geodesics> joining its points to nearby centres remain there. Squared distance $d(q,x)^2$ is then smooth in both arguments, and the <exponential map> is nonsingular on the relevant radius-$r$ tangent circles. Consequently all these distance spheres are smooth embedded one-dimensional manifolds.
Write $q=\exp_p(\varepsilon v)$ with $|v|=1$ and $\varepsilon>0$, and parameterize $S_r(p)$ by $x(\theta)=\exp_p(re(\theta))$. Define
$$
F(\varepsilon,v,\theta)=d(\exp_p(\varepsilon v),x(\theta))^2-r^2.
$$
The first variation of <geodesic> energy gives
$$
\partial_\varepsilon F(0,v,\theta)=-2r\langle v,e(\theta)\rangle.
$$
Indeed differentiating the energy of the joining <geodesic> cancels the interior term by the <geodesic> equation, leaving the initial endpoint term $-2\langle re(\theta),v\rangle$. This is also the endpoint form of the <Gauss lemma>.
Since $F(0,v,\theta)=0$, the quotient $F/\varepsilon$ extends smoothly to $\varepsilon=0$ by integrating $\partial_\varepsilon F$ from zero to $\varepsilon$. At zero it is $-2r\cos(\theta-\arg v)$, with exactly two simple zeros. The <implicit function theorem> gives two nearby simple zeros for small positive $\varepsilon$. <Compactness> of the unit vector circle makes the choice of small $\varepsilon$ uniform in $v$; away from fixed neighbourhoods of those two zeros the cosine is bounded away from zero, so there are no additional zeros.
A nonzero angular derivative at either zero says that the defining function of $S_r(q)$ has a nonzero derivative along $S_r(p)$. Thus their tangent lines differ: the intersections are transverse. Shrinking the neighbourhood of centres accordingly proves
$$
\boxed{\#(S_r(p)\cap S_r(q))=2,\qquad
\#(S_r(p)\cap S_r(q))\equiv0\pmod2
\quad(q\ne p,\ q\text{ sufficiently near }p).}
$$
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