It is not always the case that $|C \cap Hg| \le |C \cap H|$, even if G is simple. Here are two examples in small degree permutation groups, found by a brute-force search.
(1) Let $G$ be the symmetric group of degree $6$, and let $C$ be the conjugacy class of all $6$-cycles. Then $h = (1,2,3,4,5,6) \in C$ and $\mathrm{Cent}_G(h) = \left< h \right>$ contains exactly two $6$-cycles, namely $h$ and $h^{-1}$. If $g = (1,3)(2,6)$ then $\mathrm{Cent}_G(h)g$ has three $6$-cycles, namely $hg$, $h^{-1}g$ and $h^3 g$.
(2) Let $G$ be the alternating group of degree $7$, and let $C$ be the conjugacy class of elements of cycle type $(4,2,1)$. Then $h = (1,2,3,4)(5,6) \in C$ and $\mathrm{Cent}_G(h) = \left< h \right>$ contains exactly two elements of $C$, namely $h$ and $h^{-1}$. If $g = (1,5,6,7,3)$ then
$$\mathrm{Cent}_G(h)g = \lbrace (1,2)(3,4,5,7), (1,5,6,7,3), (1,4)(2,5,7,3), (2,4)(3,5,6,7)\rbrace$$
has three elements in $C$.
One small remark (related to your example): it is possible that each coset of $H$ contains a unique element of $C$. Let $G$ be a Frobenius group with cyclic kernel $K = \left< k \right>$ of prime order $p$ and complement $H = \left< h \right>$ of order dividing $p-1$. Then the conjugacy class of $h$ is $hK$. The centralizer of $h$ is $H$, so the distinct intersections in your problem are $hK \cap Hg^i = \lbrace hg^i \rbrace$, for $i \in \lbrace 0,1,\ldots,p-1\rbrace$.
A natural condition is that $G$ has finitely many connected components.
One can easily reduce this case to the connected group case, and then to the compact group case, as all maximal compact subgroups in a connected Lie group are conjugated.
Then the representation variety $\text{Hom}(\Gamma,G)$ is compact and local rigidity, that is vanishing of $H^1(\Gamma,\mathfrak{g})$, guarantees the finiteness of the number of $G$-orbits. Here $\mathfrak{g}$ denotes the Lie algebra of $G$. The fact that $H^1$ vanishes could be deduced from the fact that every isometric action of $\Gamma$ on $\mathfrak{g}$ has a fixed point, by averaging an orbit.
Best Answer
Here is a counterexample, with the Heisenberg group. Define by induction $a_0=1$ and $a_{n+1}=(a_n^2+1)a_n$. Clearly $a_n$ tends to infinity.
Let $\Gamma$ be the Heisenberg group, consisting of matrices $m(x,y,z)=\begin{pmatrix}1&x&z\\ 0&1&y\\0&0&1\end{pmatrix}$ with $x,y,z\in\mathbf{Z}^3$. Let $C_n$ be the $n$-th congruence subgroup, namely those $m(x,y,z)$ with $x,y,z\in n\mathbf{Z}$.
Let $\Gamma_n$ be the subgroup generated by the congruence subgroup $C_{a_n^2}$ and $m_n=m(1,0,a_n)$.
First check: $\Gamma_{n+1}\subset\Gamma_n$. Indeed $C_{a_{n+1}^2}\subset C_{a_n}\subset\Gamma_n$ because $a_n$ divides $a_{n+1}$. So it remains to check $m_{n+1}\in\Gamma_n$. Indeed, $m_n^{a_n^2+1}=m(a_n+1,0,a_n(a_n^2+1))=m(a_n,0,0)m_{n+1}$. Since $m(a_n,0,0)\in C_{a_n}$, we deduce $m_{n+1}\in\Gamma_n$.
Second check: $\bigcap_n\Gamma_n=\{m(0,0,0)\}$. Indeed, consider a nontrivial element $m(x,y,z)$. Choose $n$ such that $a_n$ is greater than $|x|,|y|,|z|$. Let us show that $m(x,y,z)\notin \Gamma_n$. Otherwise, it can be written as $m_n^km(aa_n^2,ba_n^2,ca_n^2)$, hence as $m(k+aa_n^2,ba_n^2,(ca_n+kba_n+k)a_n)$. Since the three coefficients are $<a_n$ in absolute value, we first deduce that $b=0$ so it equals $m(k+aa_n^2,0,(ca_n+k)a_n)$, then $k=-ca_n$, so it equals $m(a_n(aa_n-c),0,0)$, and finally it equals $m(0,0,0)$.
Finally, $m_n\in\Gamma_n$ is conjugate to the fixed element $m(1,0,0)$.