No. The reason is that
$$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr,
$$
with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives 
$$
e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right)
$$  
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$
$$
C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq  \|Q\|^4_{W^{1,4}}.
$$
Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$
$$
C_3(\|Q\|^2_{W^{1,2}} + 1) \geq  \|Q\|^4_{W^{1,4}},
$$
if you cook-up a sequence $Q_n$  with $\|Q_n\|_{W^{1,2}}=1$ and  $\|Q_n\|_{W^{1,4}}>n$  you have a contradiction.  

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form
$$
e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right).
$$
--- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that
$$
e(f,Q)\geq e(f,I)
$$
for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve 
$$
\min_{f\in \mathcal{X}} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt,
$$ 
with $\mathcal{X}=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand-> textbook question. The infimum will be in $\bar{X}$ in general, and $f$ satisfies either the Euler-Lagrange equation, or is constant. 
To find a simple answer, let us assume that $0<a<b$. Then $f>0$ in $\mathcal{X}$, and if you set $g=f^4$, you find that your problem is also
$$
\min_{g\in \mathcal{Y}} \int_a^b \left( g + \frac{1}{16} \left(\frac{dg}{dt}\right)^2\right) \,dt,
$$ 
with $\mathcal{Y}=\{f \in W^{1,2}(a,b)~:~g(a)=a^4,g(b)=b^4, g^\prime>0~a.e.\}$, a convex set. 
Now, the Euler-Lagrange equation is simply $$1+\frac{1}{8}g^{\prime\prime}=0.$$ You can solve it by hand, and playing with it, you find for example that when $$a>
\frac{\sqrt{6}}{8}\frac{\left(3554+2(33)^{3/2}\right)^{1/3}}{\sqrt{7}\left(3554+2(33)^{3/2}\right)^{2/3}+(33)^{3/2}+116\left(3554+2(33)^{3/2}\right)^{1/3}+1777)},$$ 
the solution with $g(a)=a^4$ and $g(b)=b^4$ is strictly increasing and therefore is in $\mathcal{Y}$, whereas otherwise the positivity of the gradient constraint comes into play for suitable $b$s.