The best you can do is a Lipschitz estimate at zero itself, that is, there is an $A > 0$ such that
\begin{equation*}
\sup \left\{ \|p\| \, \mid \, p \in \partial f(x), \, \, x \in B(0,r) \right\} \leq A r.
\end{equation*}
(Notice $f$ is differentiable at $0$ and $Df(0) = 0$.) In fact, there is also a lower bound:
\begin{equation*}
\|p\| \geq c_{1}\|x\| \quad \text{if} \, \, p \in \partial f(x).
\end{equation*}
The lower bound is much simpler to obtain. If $x \in \mathbb{R}^{n} \setminus \{0\}$ and $p \in \partial f(x)$, then
\begin{equation*}
0 = f(0) \geq f(x) - \langle p, x \rangle \geq c_{1} \|x\|^{2} - \langle p, x \rangle.
\end{equation*}
Hence $\langle p, \frac{x}{\|x\|} \rangle \geq c_{1}\|x\|$, implying $\|p\| \geq c_{1} \|x\|$.
To obtain the upper bound, fix $r > 0$ and assume that $\|x\| \leq r$. If $p \in \partial f(x)$ and $\xi \in S^{n-1}$, then
\begin{equation*}
c_{2} \|x + r \xi\|^{2} \geq f(x + r \xi) \geq f(x) + r \langle p, \xi \rangle \geq c_{1} \|x\|^{2} + r \langle p, \xi \rangle.
\end{equation*}
Thus,
\begin{equation*}
\langle p, \xi \rangle \leq \frac{1}{r} \left( c_{2} \|x\|^{2} + c_{2} r^{2} + 2 c_{2} r \langle x, \xi \rangle - c_{1} \|x\|^{2} \right) \leq (4c_{2} - c_{1}) r.
\end{equation*}
This gives us the desired upper bound with $A = 4c_{2} - c_{1}$.
The reason this is the best we could hope for (i.e. we can't get $Df$ Lipschitz in a neighborhood of $0$) is $c_{1} < c_{2}$. Therefore, as soon as you get away from $0$, $f$ has a positive amount of space to wiggle around --- if $f$ can wiggle around a bunch, then certainly $Df$ can't be constrained, there's no reason for it to be continuous.
To make this more precise, let's go to $n = 1$. Let $h : [0,\infty] \to [c_{1},c_{2}]$ be a non-decreasing function with a sequence of discontinuities converging at zero. Define $g : \mathbb{R} \to \mathbb{R}$ with $g(x) = -g(-x)$ by
\begin{equation*}
g(x) = 2 h(x) x.
\end{equation*}
$g$ is non-decreasing. Therefore, the function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = \int_{0}^{x} g(s) \, ds$ is convex. Furthermore, if $x > 0$, then
\begin{equation*}
c_{1}|x|^{2} \leq f(x) \leq c_{2} |x|^{2}.
\end{equation*}
Since $f(x) = f(-x)$, $f$ satisfies the desired inequalities, but $f'$ is not continuous in any neighborhood of zero.
We can generalize the previous example to $n > 1$ simpy by setting $F(x) = f(\|x\|)$.
Problem: Is there a non-convex differentiable function $f \, : \, \mathbb{R}^2 \to \mathbb{R}$ such that, for some constant $m > 0$, for all $a, b, c, d\in \mathbb{R}$,
\begin{align*}
f(a, 0) &\ge f(0, d) + a \frac{\partial f}{\partial x}(0, d) - d \frac{\partial f}{\partial y}(0, d) + \frac{m}{2}(a^2 + d^2), \tag{1}\\
f(0, b) &\ge f(c, 0) - c \frac{\partial f}{\partial x}(c, 0) + b \frac{\partial f}{\partial y}(c, 0) + \frac{m}{2}(c^2 + b^2) \tag{2} ?
\end{align*}
Yes, there is. Here is an example.
Let
$f(x, y) = x^4 + 24x^3 + 200x^2 + y^4 + 24y^3 + 200y^2$. Let $m = 8$.
It is easy to prove that $f(x, y)$ is not convex on $\mathbb{R}^2$.
The conditions are: for all $a, b, c, d\in \mathbb{R}$,
\begin{align*}
a^4 + 24a^3 + 196a^2 + 3d^4 + 48d^3 + 196d^2 &\ge 0, \\
b^4 + 24b^3 + 196b^2 + 3c^4 + 48c^3 + 196c^2 &\ge 0
\end{align*}
which are both true (easy).
Best Answer
What you want is known as Lipschitz gradient or smoothness parameter. A convex function $f$ is $\beta$-smooth if it satisfies $$ f(x) \leq f(a) + \nabla f(a)^\top (x-a) + \frac{1}{2} (x-a)^\top H (x-a) $$ for $H = \beta I$. You can see other properties of smooth function here. Your second question is related to strong convex function. Please see here for more details.