Conservation of linear momentum after explosion with friction Let's say you have a bomb at rest submerged in a liquid.  It then explodes into two equal fragments A and B which leave in opposite directions.  
If we ignore the fluid and any sound or light we can say that momentum is conserved: $p_A = -p_B$, or $p_{total} = 0$ at any time and the fragments will continue moving in opposite directions indefinitely.
Now if we add the fluid, the fragments will slowly come to a halt.  Assuming the same geometry and drag, if they slow down at the same rate, we can say that at any point in time $p_A = -p_B$, and $p_{total} = 0$ holds.   
However if our system is only the bomb fragments would it be correct to say that linear momentum is not conserved?  (since there is an external force)
 A: What we have here is an example of inconsistent use of a model and, as a result, not understanding what the complete model includes.

If we ignore the fluid

Let's remember that bit !

and any sound or light we can say that momentum is conserved: pA=−pB
  , or ptotal=0

Some explosion !  Michael Bay would be confused.  But that doesn't make any material difference to the way this works ( like most things involving Michael Bay :-) ).

at any time and the fragments will continue moving in opposite directions indefinitely.

Some kind of weird fluid which we're still ignoring.

Now if we add the fluid

Rule 1 : We can't just turn bits of our experimental model on and off when we want to.
Either the fluid is always there or it isn't.
Actually you could have "turned on" the fluid in the first place because ...

the fragments will slowly come to a halt.

.. would happen anyway and it makes no difference if you're going to turn it on now or at the start.  But that was just lucky, as you should be consistent in modeling a system and not rely on luck.

Assuming the same geometry and drag, if they slow down at the same rate, we can say that at any point in time pA=−pB
  , and ptotal=0 holds.

Yes....

However if our system is only the bomb fragments

Whoops.
Our system is not only the bomb fragments, it's the fluid too.
Remember when you said ignore the fluid.  Well now you're trying to ignore the fluid again, after telling us to pay attention to the fluid.
Rule 2 : Can't have it both ways. :-)

would it be correct to say that linear momentum is not conserved? (since there is an external force)

It would be correct to say that the fluid can slow down the fragments and in so doing absorbs energy.
Momentum is conserved because, as a whole ( fluid included ) the net momentum remains nil ( in your chosen coordinates ) and this didn't involve any external force, but it did involve an internal impulse ( the explosion ).
Note that while momentum is conserved as a whole, that doesn't mean the fluid doesn't undergo some reaction ( that energy has to go somewhere, it's just that the net momentum of the entire system is still nil ).  The fluid can take on the energy in various ways ( e.g. pressure wave, heat ).
But note, as you mention explosions, that net momentum of a system being nil does not mean the system is static and that the entire system is not e.g. spreading out to cover a larger volume.
So when you model something apply your model assumptions consistently and remember that everything in the model is part of the system and can't generally be ignored.
A: 
Now if we add the fluid, the fragments will slowly come to a halt. Assuming the same geometry and drag, if they slow down at the same rate, we can say that at any point in time =−, and =0 holds

As far as I understand , momentum cannot be conserved if there is an external force on the system.Perhaps sir, what you are overlooking here is that our system includes both the fragments and net force means vector summation of all the forces(conservative or otherwise).Provided your arguments,I can say that equal and opposite forces act on the bodies,making net force zero.

However if our system is only the bomb fragments would it be correct to say that linear momentum is not conserved? (since there is an external force)

As pointed above, there is no net external force and linear momentum is conserved.
Coming to energy arguments,they are irrelevant.It is true that both the bodies will slow down as energy is being taken from the system.But it would have no effect on net momentum provided net force is zero. That is the beauty of conservation of momentum.
A: During the explosion, that happens during a very (VERY) short time $\Delta t$, taking the system as a whole you get:
$$
\sum \mathbf{F}^{ext} = \frac{d \mathbf{p}}{dt} \approx \frac{\Delta \mathbf{p}}{\Delta t}
$$
so the change in momentum is practically:
$$
\Delta \mathbf{p} \approx \Delta t \sum \mathbf{F}^{ext}
$$
and whether we could say (total) momentum is conserved is when those external forces (drag because of the small displacement during the explosion itself, gravity etc.) are pretty small when compared to $\Delta t$, which I believe is safe to say yes, it will be conserved (for practical purposes)... after the explosion, you can keep track of the kinematics of each part using the initial speed for each part, drag, gravity and whatever...
