Your intuition is correct for a slightly different statement:
There is no function $f(x)$ on $x \ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $\epsilon > 0$ on $x>0$.
The argument is as follows: Suppose that such a function exists. Then we have $f'''(x) \geq \epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) \geq \epsilon x.
$$
Denote $f''(0) = a$, so that $f''(x) \geq \epsilon x + a$. Integrating two more times, we have
$$
f'(x) \geq \frac{1}{2} \epsilon x^2 + a x + f'(0) = \frac{1}{2} \epsilon x^2 + a x
$$ $$
f(x) \geq \frac{1}{6} \epsilon x^3 + a x^2 + f(0) = \frac{1}{6} \epsilon x^3 + \frac{1}{2} a x^2
$$
But if $f(x) < x^2$, we have
$$
x^2 > f(x) \geq \frac{1}{6} \epsilon x^3 + \frac{1}{2} a x^2
$$
for all $x > 0$, which reduces to
$$
\frac{6(1 - a/2)}{\epsilon} > x.
$$
for all $x > 0$. For any value of $\epsilon > 0$ and $a \in \mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.
Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $\frac{1}{6} x^3$ times that value. But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.