How does pH of soda drink vary with time
I will make few simplifications:


I will neglect presence of other acids like citric or phosphoric acid, considering just water and carbon dioxide.
I will use the simplified exquation of $\mathrm{pH}$ of a weak acid (  :


$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a} - \log {[\ce{CO2}]}) \tag{1}$$


I will  consider $\ce{CO2}$ escaping as the exponential process of the 1st order kinetics.


$$[\ce{CO2}]=A + B \cdot \exp{(-Ct)} \tag{1a}$$


I will  neglect the final equilibrium $\ce{CO2}$ concentration, compared to initial one:


$$[\ce{CO2}]=B \cdot \exp{(-Ct)} \tag{1b}$$

We can realize that the exponential function for $\ce{CO2}$ escaping kind of nullifies the logarithm function of $\mathrm{pH}$ definition:

$$\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {\left(B \cdot \exp{(-Ct)}\right)}\right) \tag{2}$$

$$\mathrm{pH}
\simeq \frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {B} + \frac1{2.303}Ct\right) \tag{3}$$

Therefore, $\mathrm{pH}=f(t)$ is approximately linear. 

It is not linear if we consider all factors, especially if other acids are present.



Additional requested clarification:

The equation (1) is quite notoriously known simplified equation for $\mathrm{pH}$ of weak acid. It can be directly derived from definition of a dissociation constant of an acid:

$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{4}$$

involving 2 simplifications:


solution is acidic enough to ignore water auto-dissociation:


$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{\ce{[HA]}} \tag{5}$$


solution contains enough of a weak acid, so it's majority is not dissociated:


$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{c} \tag{6}$$

where 

$$c=\ce{[HA]}+\ce{[A-]} \tag{7}$$

Then

$$\ce{[H+]}=\sqrt{ K_\mathrm{a} \cdot c} \tag{8}$$

After logaritmization of (8), we get (1).