The answer is Yes, and this is the [ultraproduct](http://en.wikipedia.org/wiki/Ultraproduct) construction. Let U be any nonprincipal ultrafilter on the set of primes. This is simply the dual filter to a maximal ideal on the set of primes, containing all finite sets of primes. (In other words, U contains the Frechet filter.) 

The quotient R/U is a field of characteristic 0. 

The ultraproduct construction is completely general, and has nothing to do with rings or fields. If M<sub>i</sub> for i in I is any collection of first order structures, and U is an ultrafilter on the subsets of I, then we may form the ultraproduct &Pi; M<sub>i</sub>/U, which is the set of equivalence classes by the relation f equiv g iff { i in I | f(i) = g(i) } in U. Similarly, the structure is imposed on the ultraproduct coordinate-wise, and this is well-defined. The most important theorem is Los's theorem, which says that &Pi; M<sub>i</sub>/U satisfies a first order statement phi([f]) if and only if { i in I | M<sub>i</sub> satisfies phi(f(i)) } in U. 

In your case, since every F<sub>p</sub> is a field, the ultraproduct is also a field. And since the set of p bigger than any fixed n is in U, then the ultraproduct will have 1+...+1 (n times) not equal to 0, for any fixed n &gt; 0. That is, the ultraproduct will have characteristic 0.

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Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F_p has the property is in the ultrafilter U. 

So you need to do some work still. What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in F<sub>p</sub>. Is this true? I'm not sure (and this may be outstripping my algebra knowledge).  This property is equivalent to asking whether every finite list of equations over Z is true in at least one F<sub>p</sub>, since one can always add one more equation so as to exclude any particular F<sub>p</sub>. If you have this for all finite collections of equations, then you can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter. 

In this case, any particular equation would have a solution in F<sub>p</sub> for a set of p in U, and os the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers. So I will let the algebraists answer whether every finite list of polynomials over Z has a solution in at least one F<sub>p</sub>. If so, the answer to your question is yes, as I have explained. If not, then the answer to your question is no, since every quotient of this ring will be an ultraproduct. 

Let me note that if one uses the product of all finite fields, instead of just the F<sub>p</sub>, then you can certainly find the desired solutions, and so the ultraproduct will be as desired.