That is an interesting problem. I think your conjecture is partially correct. After adding the $H_C=-\sum_pC_p$ term to the Hamiltonian, the ground state degeneracy will be reduced for sure. The remaining degeneracy can be 2-fold in some cases. But for the most general case, the degeneracy can be completely lifted, leaving unique ground state.
I think it would be easier to prove it using the loop algebra method developed in this work [Y.-Z. You and X.-G. Wen, arXiv:1204.0113] (which you may have already known). Admittedly, the constraint-counting method you mentioned is more rigorous, and is successful for the original toric code model. However for your case with $H_C$ term, after thinking for a while, I could not figure out how to count the constraints either.
Here is how the loop algebra works.

First, let us define 4 loop operators according to the above figure (red - $\sigma^x$, green - $\sigma^y$, blue - $\sigma^z$)
$$Q_1=\prod_{l\in\text{x-line}}\sigma_l^x, Q_2=\prod_{l\in\text{y-line}}\sigma_l^x,
P_1=\prod_{l\in\text{x-line}}\sigma_l^z, P_2=\prod_{l\in\text{y-line}}\sigma_l^z.$$
Assuming periodic boundary condition, the loops wind around the torus. Operators $Q_1$ and $Q_2$ measure the $\mathbb{Z}_2$ flux through the two torus holes respectively. While $P_1$ and $P_2$ are responsible to generate or to remove such fluxes (by moving a pair of $m$-excitations around the torus). In the original toric code model, the 4 degenerated ground states on the torus can be labeled by the eigen values of $Q_1$ and $Q_2$ (i.e. the fluxes through the holes), and different ground states are connected by the action of $P_1$ and $P_2$. Therefore the ground state Hilbert space is just the representation space of these loop operators.
It can be easily verified that these loop operators follows the algebra: $Q_1P_2=-P_2Q_1$, $Q_2P_1=-P_1Q_2$, $[Q_1,Q_2]=[P_1,P_2]=[Q_1,P_1]=[Q_2,P_2]=0$ (by checking the bound overlaps). The 4 loop operators can be divided into two anti-commuting pairs. To construct the representation for these operators, we start from $\{Q_1,P_2\}=0$. The anti-commutation relation requires $Q_1$ and $P_2$ to be represented by two Pauli matrices, say $Q_1\bumpeq\sigma_3$ and $P_2\bumpeq\sigma_1$. The same applies for $\{Q_2,P_1\}=0$. However the representation space must be enlarged such that the commutation relations between the anti-commuting pairs are also satisfied:
$$Q_1\bumpeq\sigma_3\otimes\sigma_0, Q_2\bumpeq\sigma_0\otimes\sigma_3, P_1\bumpeq\sigma_0\otimes\sigma_1, P_1\bumpeq\sigma_1\otimes\sigma_0.$$
For the original toric code model, it is easy to show that the Hamiltonian commutes with all the 4 loop operators, therefore it must act trivially in the representation space of loop operators, $H\bumpeq\sigma_0\otimes\sigma_0$. That is the algebraic reason that the ground states (and in fact each energy level) must be (at least) 4-fold degenerated.
Now let us consider the effect of adding $H_C$ term. All the $C_p$ operators can be divided into two classes: those away from $Q_{1,2}$ loops as $C_{p1}$ in the figure, and those adjacent to $Q_{1,2}$ loops as $C_{p2}$ in the figure. The Hamiltonian is the sum of these two classes of $C_p$ operators: $H_C=H_{1}+H_{2}$, with $H_{1,2}=-\sum C_{p1,p2}$. Note that $H_1$ commutes with all the loop operators still. However $H_2$ anti-commutes with $Q_1$ or $Q_2$ while commutes with $P_1$ and $P_2$. Therefore $H_1$ and $H_2$ are represented differently: $H_1\bumpeq\sigma_0\otimes\sigma_0$, $H_2\bumpeq a\sigma_1\otimes\sigma_0+ b\sigma_0\otimes\sigma_1+ c\sigma_1\otimes\sigma_1$ (combined with some coefficients $a,b,c$ to be determined case by case). $H_2$ induces mixing between the degenerated ground states. According to the idea of degenerate perturbation, the degeneracy will be lifted, and in the worst case, completely lifted.