No! For a particle in a box with rigid boundary conditions (i.e., the box with infinitely high potential walls, which already shows that this is an artificial example to begin with) there doesn't even exist a momentum observable.
This can be easily shown as follows: The momentum operator is, by definition via Noether's theorem, the operator that generates spatial translations. Let's look at the 1-dimensional box for simplicity. From this definition of momentum it follows that
\hat{p}=-\mathrm{i} \partial_x.
The eigenvalue problem reads
\hat{p} u_p(x)=-\mathrm{i} \partial_x u_p(x)=p u_p(x).
It follows that
u_p(x)=N(p) \exp(\mathrm{i} p x).
However, the Hilbert space of wave functions of the finite box is given by the functions that vanish at the boundaries of the box, e.g., the interval (-L/2,L/2).
This boundary condition is not fulfilled by any of the putative eigenfunctions and thus there is no essentially self-adjoint momentum operator on this Hilbert space, and thus there is no momentum observable in the usual sense.
The whole thing is different for the Hamiltonian, which you formally define as
\hat{H}=\frac{\hat{p}^2}{2m}=-\frac{1}{2m} \partial_x^2.
The eigenfunctions are
\hat{H} u_E(x)=-\frac{1}{2m} \partial_x^2 u_E(x)=E u_E(x).
This leads to the solutions
u_E(x)=N(E) [\exp(\mathrm{i} k x)+B \exp(-\mathrm{i} k x)],
where k=\sqrt{2m E}.
Now we can fulfill the boundary conditions:
\exp(\mathrm{i} k L/2)+B \exp(-\mathrm{i} k L/2)=0,
\exp(-\mathrm{i} k L/2) + B \exp(\mathrm{i} k L/2)=0.
From the first equation you get
B=\exp(2 \mathrm{i} k L/2)=\exp(\mathrm{i} k L)
and from the second
B=\exp(-\mathrm{i} k L)
This means
\exp(2\mathrm{i} k L)=1.
This determines the possible values of k by the condition
2 k L=2 \pi n \; \Rightarrow \; k=\pi n/L,
where n \in \mathbb{Z}.
Now for each n \in \mathbb{Z} you can solve for B using one of the equations. This gives
B_n=(-1)^n.
Thus you have
u_E(x)=N(E) [\exp(\mathrm{i} \pi n x/L)+(-1)^n \exp(-\mathrm{i} \pi n x/L)].
Now switching from a given n \in \mathbb{N} to -n leads to the same solution (up to a sign, which is insignificant, because it's just a phase factor). Thus you have to take only n \in \{0,1,2,3,\ldots\}.
These solutions are a complete orthogonal set in L^2[(-L/2,L/2)], because it's just
u_n(x)=\begin{cases}<br />
N_n \cos(\pi n x/L) & \text{for} \quad n \quad \text{even},\\<br />
N_n \sin(\pi n x/L) & \text{for} \quad n \quad \text{odd}.<br />
\end{cases}<br />
The normalization constant is given by
\int_{-L/2}^{L/2} |u_n(x)|^2=1 \; \Rightarrow \;N_0=1/\sqrt{L}, \quad N_n=\sqrt{2}{L} \quad \text{for} \quad n \in \{1,2,\ldots\}.
The well-known theorems about Fourier series thus tell you that indeed this is a complete set of orthonormal function, because obviously
\langle u_n|u_{n'} \rangle=\delta_{nn'}
with the so chosen normlization constants.
