Yes. This follows from the classical uniqueness theorem due to Osgood.

> **Osgood's Criterion.** Let $\omega(t,u)=\phi(t)\psi(u)$ where $\phi(t)\geq 0$ is continuous 
on the interval $(0,a)$ and $\psi(u)$ is continuous on $\mathbb R_{+}$, $\psi(0)=0$, $\psi(u)>0$ for $u>0$, and 
$$\int_{0}^{\epsilon}\phi(t)dt<\infty,\qquad \int_{0}^{\epsilon}\frac{du}{\psi (u)}=\infty$$
for some $\epsilon>0$. Suppose that the function $f:[0,a]\times B_R(x_0)\to \mathbb R^d$ satisfies the condition
$$||f(t,x_1)-f(t,x_2)||\leq\omega (t,||x_1-x_2||)$$
for any $t\in(0,a]$ and $x_1,x_2\in B_R(x_0)$. Then the initial value problem
$$\dot{x}=f(t,x),\qquad x(0)=x_0$$
has at most  one solution on the interval $[0,\delta]$ with some $\delta>0$.

Osgood's theorem allows for functions $f(t,x)$ which are discontinuous at $t=0$ (in fact, the condition that $\phi(t)$ is continuous can be replaced with a much weaker  condition
of integrability).  Of course, the existence of a local solution is implied by the Peano theorem under the additional assumption that $f$ is continuous in $(t,x)$.