Your question is muddled and unclear (in my professional opinion - you may think otherwise and that's OK), but I think I can make one thing clearer.
- If you have two entangled systems, $A$ and $B$, and perform independent experiments on either side, then nothing about the choice of experiment you perform on $B$ will have any effect on the local results from $A$.
Note the keywords independent and local. If you look for correlations between the two experiments, then the situation can change. Thus, if you post-select the results from $A$ to include only the runs when $B$ performed experiment $B_1$ and obtained outcome $B_1(+)$, then the post-selected results from $A$ will depend, in general, on both the measurement and the outcome.
Having said this, I will continue to hammer on on what's been said before: your question is essentially undecidable unless you specify exactly what type of entanglement you imagine both photons to share. You seem to have a magical view of entanglement (which is alarmingly common) in which touching system $B$ will immediately and irrevocably change everything about system $A$, which is very, very far from the truth.
Allow me to elaborate by presenting two scenarios, which I believe are consistent with how you phrased your question:
Scenario 1
The photons are entangled in polarization, sharing the state $|HH\rangle+|VV\rangle$, but are otherwise indistinguishable.
In this case, nothing you do on $B$ will have any effect on the interference pattern (or lack thereof) shown in $A$, because the entanglement simply does not couple to the spatial modes.
Scenario 2
The photons are spatially entangled, in such a way that when photon $A$ hits its left slit then photon $B$ hits its right slit, and vice versa, i.e. they share the state $|LR\rangle+|RL\rangle$.
In this case, your actions on $B$, both your choice of measurement and the corresponding experimental outcome, can affect the results of $A$ if you postselect on the appropriate results on $B$, at least in principle. Other choices of measurement in $B$ will not have any effect on $A$. In particular:
If you perform a path measurement on $B$, where you detect which-way information for it, then (in this scenario) you also make available which-way information on $A$, which precludes it from showing an interference pattern. This is independent of the temporal sequence and spacing between the two measurements, and on whether the measurement in $B$ was randomized or not.
Suppose, on the other hand, that you do nothing with the photon in $B$ and allow it to run through the slits undisturbed. In this case (for this scenario), you will not observe any interference pattern on either photon. The reason for this is that each photon contains (in principle) which-way information on the other, so that the other one cannot interfere. The reduced state for both is $$\rho=\frac12\left(|L\rangle\langle L|+|R\rangle\langle R|\right),$$ and no coherent dynamics can be extracted from it. This is independent of the temporal sequence and spacing between the two measurements, and on whether the measurement in $B$ was randomized or not.
On the other hand, within this scenario there is indeed one measurement choice on $B$ which can restore the interference pattern on $A$, and it is of course the quantum eraser scheme. If you measure in the basis $\{|+\rangle,|-\rangle\}=\{|L\rangle+|R\rangle,|L\rangle-|R\rangle\}$, then the detector on $A$ will not turn up any interference pattern, but if you post-select and separate the counts on $A$ which coincided with $|+\rangle$ and $|-\rangle$ detections on $B$, then the blob will separate into two complementary interference patterns.
This behaviour is independent of the temporal sequence between the two measurements, their temporal separation, and whether the measurement in $B$ was randomized or not. If the measurement in $A$ is performed before the randomized decision to measure in $B$, then you have the delayed-choice quantum eraser experiment, which has been analyzed in detail elsewhere and which I will not attempt to clarify.
Note also that in the latter case, there is no measurement at $A$ that will enable you to predict the outcome in $B$. This is, again, exactly the same as with the delayed-choice quantum eraser, regardless of however many photons you send.
These two scenarios demonstrate that your question is ill-defined - the precise answer depends on information not contained in the question.
Some additional notes:
The measurement outcome is independent of however many photons you send through. Even in case 3 of scenario 2, if you try to send a million photons and observe the screen before you measure (or decide the measurement basis, or toss your coin, or whatever) on $B$, what you'll see in the screen in $A$ is... nothing, just an interferenceless blob. This is because each photon is an independent run of the experiment, and will have an independent result on the $|\pm\rangle$ basis, so if you add them up you are already performing a decision of what you'll do with the results from $B$.
The need for many photons to detect an interference pattern is a red herring; the "objection of many" is only raised by people with a hazy understanding of the quantum eraser experiment.
The goal is to detect coherence between $|L\rangle$ and $|R\rangle$ on the $A$ side, and this can be done without the need for a large number of photons, by being more clever about it. More specifically, you should collect the light behind each slit with an optical fiber and then shine them onto a beam splitter with detectors $M$ and $N$ on the output ports, calibrated so that $|L\rangle+|R\rangle$ will go exclusively into $M$ and $|L\rangle-|R\rangle$ will go exclusively into $N$. Thus a single count in $M$ rules out $|-\rangle$ and a single count in $N$ rules out $|+\rangle$.
This scheme is (provably) optimal. It will, of course, not give you enough information to completely determine the state from a single run of the experiment; this, however, is a fundamental limitation and the only way to get around it is via quantum state tomography.
This answer represents my best attempt in good faith to make sense of your question, which is a very tall proposition. If you feel I have not understood any aspect of the question, I would encourage you to think long and hard about how you are phrasing your post, instead of chucking more rep at it. Science depends crucially on its communication, and this includes making sure that one's audience can understand the material. If the entire audience says it is too obscure, blaming the audience will not benefit you at all; instead, try to clarify the material, and politely ask your audience what else you could do to make the point more evident. (If you want to, of course. Blaming the audience is lots of fun too.)
I'd like to expand my earlier comment into a little essay on the severe practical difficulties in performing the suggested experiment.
I'm going to start my asserting that we don't care if the experiment is a "two-slit" per se. It is sufficient that it is a diffractive scattering experiment of some kind.
However, we do care about having
spacial resolution good enough to distinguish which scattering site (or slit) was the one on the path of the alleged particle
the ability to run the experiment at low rate so that we can exclude multi-projectile or beam/beam interaction as the source of any interference that we observe. (Though it's going to turn out that we never even get far enough for this to matter...)
Now let's get down to designing the beast.
To start with we should note to any casual readers that the diagrams you see in pop-sci treatment are not even remotely to scale: typical classroom demonstration kit for use with lasers has the slits set less than $1\,\mathrm{mm}$ apart and uses projection distances of several meters or more to get fringes that are separated by a few centimeters. Or then use much closer set slits to get large angles.
The angular separation between maxima is on order of
$$ \Delta \theta = \frac{\lambda}{d} \,,$$
where $\lambda$ is the relevant wavelength and $d$ is the scattering site (or slit) separation. Allowing that the distance from the scattering surface to the projection surface is $\ell$, the spacial separation is (in the small angle approximation)
$$ \Delta x = l \Delta \theta = \frac{\ell}{d} \lambda \,.$$
Anna has suggested doing the experiment with electrons, which means that we're interested in the de Broglie wavelength usually given by $\lambda = \hbar/p$, and measuring their position en route with a tracking detector of some kind.
The tracking detector's spacial resolution is going to be the big barrier here.
Let's start by considering a Liquid Argon TPC because it is a hot technology just now. Spacial resolution down to about $1 \,\mathrm{mm}$ should be achievable without any breakthrough in technology (typical devices have $3$-$5\,\mathrm{mm}$ resolution). That sets our value for $d$.
Now, to observe a interferences pattern, we need a detector resolution at least four times finer than the spacial resolution.
Assume for the sake of argument that I use a detector with a $20 \,\mathrm{\mu{}m}$ spacial resolution. Maybe a MCP or a silicon tracker. That sets $\Delta x = 4(20 \,\mathrm{\mu{}m})$.
I also assume that I need $\ell$ to be at least $2d$ to be able to track the particle between the scattering and projection planes. Probably an under-estimate, so be it. Now I can compute the properties of the necessary electron source
$$\begin{align*}
p &= \frac{\hbar}{\lambda} \\
&= \frac{\hbar\ell}{d \, \Delta x} \tag{1}\\
&= 2\frac{\hbar}{\Delta x}\\
&= \frac{7 \times 10^{-22} \,\mathrm{MeV \, s}}{40 \times 10^{-6} \,\mathrm{m}}\\
&= \frac{7 \times 10^{-22} \,\mathrm{MeV}}{7 \times 10^{-12} c} \\
&= 10^{-10} \,\mathrm{MeV/c}\\
&= 10^{-4} \,\mathrm{eV/c} \,,
\end{align*}$$
which is safely non-relativistic, so we have a beam energy of $5 \times 10^{-9}\,\mathrm{eV^2}/(m_e c^2)$, and the tracking medium will completely mess up the experiment.
By choosing a $20\,\mathrm{m}$ flight path between scattering and detection and
getting down to, say, the $10\,\mathrm{\mu{}m}$ scale for $d$ we can get beam momenta up to $10^3\,\mathrm{eV}$ which at lest gives us beam energies about $1\,\mathrm{eV}$. But how are you going to track a $1\,\mathrm{eV}$ electron without scattering it?
I'm sure you can get better spacial resolution in silicon, but I don't think you can get the beam energy up high enough to pass a great enough distance through the tracking medium to actually make the measurement.
The fundamental problem here is the tension between the desire to track the electron on it's route which forces you to use nearly human scales for parts of the detector and the presence of that pesky $\hbar$ in the numerator of equation (1) which is driving the necessary beam momentum down.
The usual method of getting diffractive effects is just to make $d$ small and $\ell$ large enough to compensate for the $\hbar$ but our desire to track the particles works against us there by putting a floor on our attemtps to shrink $d$ and by because longer flight paths mean more sensitivity to scattering by the tracking medium.
Best Answer
I think the experiment you are proposing is not possible in the way you want it.
Let us say we produce two photons in an electron-positron-annihilation with total momentum zero. (Since I don't see an easy way to produce entangled electrons I will talk about photons here, but I think it is not important for the argument). Those two photons are of course entangled in momentum: if one has momentum $\vec p$ the other one has momentum $-\vec p$.
But in order to make this statement you have to make a moemntum measurement on the initial state, i.e. know that the total momentum is zero with a certain $\Delta p$. But then, by means of the uncertainty relation, you only know the position where the photons were emitted with an uncertainty $\Delta x \propto (\Delta p)^{-1}$.
Now you can have two scenarios: Either your double-slit is small enough and far enough away that due to the uncertainties $\Delta p$ and $\Delta x$ you do not know through which slit your photon goes. Or you still can tell (with some certainty).
In the second case there will never be an interference pattern. So no need for entanglement to destroy it.
But in the first case, due to the uncertainty $\Delta x$, measuring the position (by determining which slit your photon takes) does not give you an answer about the entangled photons position that is certain enough to tell which slit it will go through. Therefore you will see interference on both sides.
So an EPR like measurement is not possible in the experimental setup you propose. I would assume that in general you need commuting observables, like spin and position in the Stern-Gerlach experiment, in order to measure EPR. But I didn't think that through yet.
addendum, 03-19-2014:
Forget about the second photon for a while. The first photon starts in a position state which is a Gaussian distribution around $\vec x_0$ and a momentum state which is a Gaussian around $\vec p_0$. After some time $t$ its position has evolved into a Gaussian of $\mu$ times the width around $\vec x_0 + \vec p_0 t$ (mass set equal to 1) while the momentum state is now $1/\mu$ times the width around $\vec p_0$. So while your spatial superposition gets larger - and thus better to measure with a double slit - the superposition in the momentum state, in which you have entanglement, gets smaller. You don't gain anything from entanglement, since your momentum wave-function is so narrow, that you know the momentum anyways.
It is actually not important to have space and momentum for this. Just take any non-commuting observables A and B, say with eigenstates A+, A-, B+, B-, and take two states S1 and S2 that are entangled in A. So measuring S1 in A+ implies S2 in A- and vice versa. But what you want is measure if S1 is in B+ or B- and from this conclude if S2 is in B+ or B-. And since A and B do not commute, measuring B with some certainty gives you a high uncertainty on A, meaning, for knowing if S1 is in B+ or B- you completely loose the information if it is in A+ or A-. So you cannot say anything about S2. On the other hand, as long as you are still in an eigenstate of A and know what to expect for the A measurement of S2, you don't know anything about the result of the B measurement.
So in order to do an EPR experiment you need entanglement in the observable you measure or an observable that commutes with it.
Please tell me if my thoughts are wrong.