The difference is

$$X_t^1-X_t^2=\int_0^t\, \mu_1(X_s^1)-\mu_2(X_s^2)ds + \int_0^t\,\sigma(X_s^1)-\sigma(X_s^2)dW_s.$$

For the second term, you can use that it is a martingale and apply Burkholder-Davis-Gundy (see Revuz-Yor 2.4 theorem in SDE section) to bound

$$E[\sup_{t\leq T}|\int_0^t\,\sigma(X_s^1)-\sigma(X_s^2)dW_s|]$$

$$\leq E[(\int^{T} |\sigma(X_s^1)-\sigma(X_s^2)|^{2}ds)^{1/2}]$$

$$\leq L_{\sigma}E[(\int^{T} |X_s^1-X_s^2|^{2}ds)^{1/2}]$$

$$\leq L_{\sigma} T^{1/2} E[\sup_{s\leq T}|X_s^1-X_s^2|]$$

for $L_{\sigma}$ the Lipschitz constant for $\sigma$. So if $L_{\sigma} T^{1/2}<1$ we can bound

$$E[\sup_{s\leq T}|X_s^1-X_s^2|]\leq \frac{1}{1-L_{\sigma} T^{1/2}}E[\sup_{s\leq T}|\int_0^t\, \mu_1(X_s^1)-\mu_2(X_s^2)ds|]$$

$$\leq \frac{1}{1-L_{\sigma} T^{1/2}}E[\int_0^T\, |\mu_1(X_s^1)-\mu_2(X_s^2)|ds].$$

Now we need to add and subtract $\mu_{2}(X_s^1)$ and use that its $L_{\mu_{2}}$-Lipschitz to bound

$$\leq \frac{1}{1-L_{\sigma} T^{1/2}}E[\int_0^T\, |\mu_1(X_s^1)-\mu_2(X_s^1)|ds]+\frac{TL_{\mu_{2}}}{1-L_{\sigma} T^{1/2}}E[\sup_{s\leq T}|X_s^1-X_s^2|].$$

So again if $\frac{TL_{\mu_{2}}}{1-L_{\sigma} T^{1/2}}<1$ we bound

$$E[\sup_{s\leq T}|X_s^1-X_s^2|]\leq c_{1}c_{2}T\|\mu_{1}-\mu_{2}\|_{\infty},$$

for
$$c_{1}:=\frac{1}{1-L_{\sigma} T^{1/2}},c_{2}:=(1-\frac{TL_{\mu_{2}}}{1-L_{\sigma} T^{1/2}})^{-1}.$$

Just as a side note, if you have some comparison $\mu_{1}\leq \mu_{2}$, then you can get comparison for the solutions eg. see "On pathwise uniqueness and comparison of solutions of one-dimensional stochastic differential equations".