If there were such a functional $\mathcal{F}$, observe that

Under Ricci flow the functional would have to decrease. That is, if $\partial_t g(t) = -2 Rc[g]$ then $\partial_t \mathcal{F}(g(t)) \leq 0$, with strict inequality of $-2 Rc[g] \neq 0$.

The functional would have to be invariant under diffeomorphisms, that is if $\Phi$ is a diffeomorphism of the manifold then $\mathcal{F}(\Phi^*g) = \mathcal{F}(g)$. This is because the Ricci flow is invariant under diffeomorphism.

The point is that Robert Bryant found a metric on $\mathbb R^n$ which moves under Ricci flow only by diffeomorphisms (which is called a steady soliton). That is, there is a metric $g_S$ and a family of diffeomorphisms $\Phi_t$ such that $g(t) = \Phi^*_t g_S$ is the solution of Ricci flow starting from $g_S$. By point 2 above, $\mathcal{F}(g(t)) = \mathcal{F}(g_S)$, which contradicts point 1.

The details of the constructing the soliton are written by Robert Bryant here. Many other gradient steady solitons have been discovered since then.

**Edit**: As pointed out by Rbega, this argument only shows that such a functional would have to be infinite on the Bryant soliton. We can say that mean curvature flow is the gradient flow for area. The gradient of area is defined even for surface with infinite area, by consider compact perturbations. Therefore, we need a stronger argument.

If there was a compact steady soliton, then the argument above would work (assuming that $\mathcal{F}$ is always finite on compact manifolds). However, there is no compact steady soliton (except for Ricci flat manifolds).

The stronger argument comes from the existence of shrinking solitons on compact manifolds. The first known example is the Koiso-Cao soliton. A shrinking soliton is a pair of a metric $g$ and a vector field $X$ which solves
$$-2Rc = - g + \mathcal{L}_X g,$$
equivalently
$$-2Rc_{ij} = - g _{ij}+ \nabla_i X_j + \nabla_j X_i.$$
Under Ricci flow, a steady soliton shrinks and also move by diffeomorphisms integrating $X$.

If $-2Rc$ where the gradient of $\mathcal{F}$ then,
$$-Grad \mathcal{F} = - g + \mathcal{L}_X g$$
I claim this can't be true unless $\mathcal{L}_X g = 0$. Note that the two parts of this tensor field are orthogonal in $L^2$, because
$$\int_{M} (g,\mathcal{L}_X g)_g dVol_g = \int_M tr(\mathcal{L}_X g) dVol_g = \int_M div_g(X) dVol_g = 0.$$
Therefore we have
\begin{align}
|\mathcal{L}_X g|^2_{L^2}
&= (-Grad \mathcal{F}, \mathcal{L}_X g)_{L^2} \\
&=-\partial_{\epsilon} \mathcal{F}(g + \epsilon \mathcal{L}_X g) \\
&= -\partial_{\epsilon} \mathcal{F}(\Phi_{\epsilon}^* g) = 0
\end{align}
The first equality is by the orthogonality of $-g$ and $\mathcal{L}_X g$, the second is the defining property of the gradient.