Solution: Note that \begin{align*} f^{\prime\prime\prime} (x) = 1 & \implies f^{\prime\prime} = x + a \\ & \implies f^\prime (x) = \frac{x^2}{2} + ax + b \\ & \implies f(x) = \frac{x^3}{6} + \frac{ax^2}{2} + bx + c \\ & \kern 1.65cm = \frac{x^3}{6} + Ax^2 + Bx+C, \end{align*} where $A,B,C$ are constants. Given that $p(x)$ is the quadratic polynomial which interpolates $f(x)$ at $0,2$ and $3$, so write \[ p(x) = \tilde{A} x^2 + \tilde{B} x + \tilde{C}. \]

Now we have \begin{align*} p(0) = f(0) & \implies \tilde{C} = C \\ p(2) = f(2) & \implies 4\tilde{A} + 2\tilde{B} + \tilde{C} = \frac{4}{3} + 4A + 2B +C \\ p(3) = f(3) & \implies 9\tilde{A} + 3\tilde{B} + \tilde{C} = \frac{9}{2} + 9A + 3B +C \end{align*} Solving the above, we will get \[ A-\tilde{A} = -\frac{5}{6},~B-\tilde{B} = 1,~ C-\tilde{C} =0. \]

So, \begin{align*} p(1) - f(1) & = \left( \tilde{A} + \tilde{B} + \tilde{C} \right) - \left( \frac{1}{6} + A + B + C \right) \\ & = \left( \tilde{A}-A \right) + \left( \tilde{B} -B \right) + \left( \tilde{C} - C \right) - \frac{1}{6} \\ & = \frac{5}{6} + (-1) + 0 - \frac{1}{6} \\ & = -\frac{2}{6} = -\frac{1}{3}. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{p(1)-f(1) = -\frac{1}{3}}} \]