Yes, it trivially happens if $G = \langle g \rangle$ is the cyclic group of order $pq$ -- then all $3$ conjugacy classes mentioned have precisely one element. A smallest example of a nonabelian group $G$ having such an element is $$ G \ = \ \langle (1,3,2)(4,5)(6,7), (2,3)(5,7) \rangle, \ \ g \ = \ (1,2,3)(4,5)(6,7). $$ Then $|G| = 24$, and we have $|(g^2)^G| = |(g^3)^G| = |(g^{3-2})^G| = 2$.

Yes, it trivially happens if $G = \langle g \rangle$ is the cyclic group of order $pq$ -- then all $3$ conjugacy classes mentioned have precisely one element. A smallest example of a nonabelian group $G$ having such an element is $$ G \ = \ \langle (1,3,2)(4,5)(6,7), (2,3)(5,7) \rangle, \ \ g \ = \ (1,2,3)(4,5)(6,7). $$ Then $|G| = 24$, and we have $|(g^2)^G| = |(g^3)^G| = |(g^{3-2})^G| = 2$.

A nontrivial example where $p \neq 2$ is the following group of order $675$ and structure $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm C}_9) \rtimes {\rm C}_3$:

G = < (1,2,6,4,9,8,5,7,3)(10,15,23)(12,17,24)(13,18,22)(14,16,25)(19,29,28)
      (20,21,32)(26,27,33)(30,34,31)(35,40,48)(37,42,49)(38,43,47)(39,41,50)
      (44,54,53)(45,46,57)(51,52,58)(55,59,56)(60,65,73)(62,67,74)(63,68,72)
      (64,66,75)(69,79,78)(70,71,82)(76,77,83)(80,84,81), 
      (2,7,9)(3,6,8)(10,41,73,32,56,60,16,48,82,31,35,66,23,57,81)
      (11,45,79,33,59,61,20,54,83,34,36,70,29,58,84)   
      (12,46,80,24,43,62,21,55,74,18,37,71,30,49,68)
      (13,47,78,15,52,63,22,53,65,27,38,72,28,40,77) 
      (14,44,67,26,50,64,19,42,76,25,39,69,17,51,75) >

-- when choosing the element

g = (2,7,9)(3,6,8)(10,41,73,32,56,60,16,48,82,31,35,66,23,57,81)
    (11,45,79,33,59,61,20,54,83,34,36,70,29,58,84)
    (12,46,80,24,43,62,21,55,74,18,37,71,30,49,68)
    (13,47,78,15,52,63,22,53,65,27,38,72,28,40,77)
    (14,44,67,26,50,64,19,42,76,25,39,69,17,51,75),

we have $|(g^3)^G| = |(g^5)^G| = |(g^{5-3})^G| = 3$.