There are two different issues. One of them is the sign of the momentum; the other one is whether the momentum is spread (it's not because of the unnatural boundary conditions).
Concerning the first point, the standing wave (sine) is a real function and every real wave function has the same probability to carry momentum $+p$ and $-p$. So indeed, both of them are equally likely.
But even if you write the sine as a difference of two complex exponentials, it's still true that these exponentials aren't equal to the wave function everywhere – just inside the box – so it's still untrue that the momentum is sharply confined to two values $p$ and $-p$.
To get the probabilities of different momenta, you need to Fourier transform the standing wave – a few waves of the sine. One has
$$\int_0^1 dx\,\sin(n\pi x)\exp(ipx) = \frac{n\pi[-1 +e^{ip}(-1)^n]}{p^2-n^2\pi^2} $$
Square the absolute value to get the probability density that the momentum is $p$. The momentum $p$ should have the natural prefactors $\hbar/L$ etc. and the overall wave function should get another normalization factor for the overall probability to equal one. This changes nothing about the shape of the probability distribution: almost all values of $p$ have a nonzero probability.
$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse.
Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the $z$-direction $\ket{\uparrow_z}$,$\ket{\downarrow_z}$.
Now, the state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{\uparrow_z} + \ket{\downarrow_z})$ is not an eigenstate of $S_z$, and measurement of $S_z$ will collapse it with equal probability into $\ket{\uparrow_z}$ or $\ket{\downarrow_z}$. Yet, this is an eigenstate of the spin in $y$-direction, i.e. $\ket{\psi} = \ket{\uparrow_y}$ (or down, we'd have to check that by computation, but it doesn't matter for this argument). So, although you can "collapse" $\ket{\psi}$ into other states, it already looks like a collapsed state by your logic. This shows that the notion of "looking like a collapsed state" is not very useful to begin with.
Furthermore, you seem to be confused about the difference between a measurement (inducing collapse in some dictions) and the application of an operator. You say
So when the hamiltonian acts on a system in one of those eigenfunctions, that system always collapses to the same point in space?
but this is non-sensical. The action of the Hamiltonian is an infinitesimal time step, as the Schrödinger equation tells you:
$$ \mathrm{i}\hbar\partial_t \ket{\psi} = H\ket{\psi} $$
and, for an eigenstate $\ket{\psi_n}$, which is a solution to the time-independent equation with energy $E_n$, you have by definition $H\ket{\psi_n} = E_n\ket{\psi_n}$, that is, the Hamiltonian is a "do nothing" operation on eigenstates since multiplication by a number does not change the quantum state. That is, after all, why we are interested in the solutions to the time-independent equation - because these are the stationary states that do not evolve in time. This has nothing to do with collapse, or measurement.
Lastly, exactly determinate states of position are not, strictly speaking, quantum states, since the "eigenfunctions" of the position operator "multiplication by x" are Dirac deltas $\psi(x) = \delta(x-x_0)$, which are not proper square-integrable functions $L^2(\mathbb{R})$ as quantum states are usually required to be. But yes, this is "a spike", and conversely, the determinate momentum states are plane waves $\psi(x) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar}px}$.
Best Answer
No.
$\hat p\sin(kx)$ is not a multiple of itself. It is true that $\hat p\,e^{\pm i k x}= \pm\hbar k e^{\pm i k x}$ but a (complex) linear combination of two eigenfunctions is only an eigenfunction if both eigenfunctions in the sum have the same eigenvalue, which is NOT the case here as the eigenvalues $\pm \hbar k$ differ by a sign.
$\sin(kx)$ is an eigenfunction of $\hat p^2$ (although be careful as it is not a normalizable eigenfunction in the sense that $\int_{-\infty}^\infty \sin^2(kx) dx$ is not finite).
In response to comments: Let $\hat T$ be your favorite operator, and let $\phi_A(x)$ and $\phi_B(x)$ be such that $$ \hat T\phi_A(x)=A\phi_A(x)\, ,\qquad \hat T\phi_B(x)=B\phi_B(x)\, . $$ Take the complex linear combination $$ \psi(x)=\alpha \phi_A(x)+ \beta \phi_B(x) $$ and look at \begin{align} \hat T\psi(x)&= A\alpha \psi_A(x)+B\beta\phi_B(x)\\ &=A(\alpha \psi_A(x)+\beta\phi_B(x))+(B-A)\beta \phi_B(x)\\ &=A\psi(x)+(B-A)\beta\phi_B(x) \end{align} For this to be a multiple of the original $\psi(x)$ one must "cancel" the extra $\phi_B(x)$ term and so have $A=B$, i.e. the eigenvalues are the same, or $\beta=0$, meaning you did not have a linear combination to start with.
The calculation of average values is done using \begin{equation} \int dx\, \psi^*(x) \left[ \hat T\psi(x)\right]\, \end{equation} You can use the explicit expression of $\psi(x)$ above and find, if your functions $\phi_A(x)$ and $\phi_B(x)$ are orthonormal in the sense that $$ \int dx \phi_i^*(x) \phi_j(x)=\delta_{ij} $$ that the average value of $\hat T$ comes out as $\alpha^*\alpha A+\beta^*\beta B$. Of course if $A=B$ and $\alpha^*\alpha +\beta^*\beta=1$ then the average value is just the eigenvalue, but this is really a special case of the general expression for the average value. The average value makes sense as the integral above for any $\psi(x)$, irrespective of whether or not one know the expansion of $\psi(x)$ in terms of eigenfunctions of $\hat T$.
In the case of $\sin(n\pi x/L)$ and $\hat p$, the mathematical machinery can bypassed by a physical argument. The quantity $$ \psi^*(x)\left[\hat p \psi(x)\right]= i\hbar (n\pi/L) \sin(n\pi x/L)\cos(n \pi x/L) $$ is purely imaginary, so the integral will also be purely imaginary. However, $\hat p$ is an observable so its average values must be real. This is a contradiction unless the average value is $0$. This "trick" will apply whenever the wavefunction is real and the observable is represented by an operator of the form $i\times$(something real). Another situation where this trick would be useful would be the average value of $\hat p$ for a wave function that is a real linear combination of harmonic oscillator states.