Mathematica

 Print[""<>IntegerString[MaximalBy[Tally[Join@@IntegerDigits[Prime[Range[PrimePi[#1]+1,PrimePi[#2]]]]],Last][[;;,1]]&@@FromDigits/@StringSplit@InputString[]]]

Time Specs

Using Mathematica's builtin timer, AbsoluteTiming, I got these times for each listed range.

f=MaximalBy[Tally[Join@@IntegerDigits[Prime[Range[PrimePi[#1]+1,PrimePi[#2]]]]],Last][[;;,1]]&
Grid[Table[{1,10^x+1,AbsoluteTiming[f[1,10^x+1]][[1]]},{x,1,7}]]

Returns:

1        11   0.0003388
1       101   0.0003645
1      1001   0.0016222
1     10001   0.0040902
1    100001   0.0267737
1   1000001   0.211297
1  10000001   4.55387
1 100000001  43.9889

It can go over a range of a hundred million numbers in under a minute.

Usage

wolframscript -f primedigitcount.wls
? 21 41
3
wolframscript -f primedigitcount.wls
? 51 89
7
wolframscript -f primedigitcount.wls
? 201 501
3

Ungolfed

Takes the space-seperated input and converts it to two integers

a = FromDigits/@StringSplit@InputString[]

PrimePi is the inverse function of Prime, so this gets all the primes between the two a's. The +1 is that if the operand is not prime, it gets the first prime below it, but we want a prime above the smaller number.

b = Prime[Range[PrimePi[#1]+1,PrimePi[#2]]&@@a

Turns every number in the list into a list of it's digits, and then concatenates all the lists into one list of all the digits

c = Join@@IntegerDigits[b]

Mathematica has a nice built-in for turning a list into unique items and their counts.

d = Tally[c]

Takes a list of the item (or items) who has the largest count, and strips off the tally.

e = MaximalBy[d,Last][[;;,1]]

Converts every number in the list to a string, concatenates them, and prints it.

Print[""<>IntegerString[e]]

Conjecture:

In hope that somebody more mathematically inclined could find this to be an O(n) problem, I leave you this conjecture:

If the first number is kept as 1, and the 2nd number tends to be large (n>100), the result will be no more than two of 1, 3, or 7.

Python 3

import collections
p=lambda n:n>1 and all(n%i for i in range(int(n**0.5),1,-1))
print(collections.Counter(''.join(map(str,filter(p,range(*map(int,raw_input().split(' '))))))).most_common(1)[0][0])

Probably not the best way it can be done, but a start. (196 bytes)

Time Specs

Using Python's builtin code timer module, timeit, I got these times for each listed range.

(1,       11):      0.000084381103520
(1,      101):      0.000338881015778
(1,     1001):      0.003911740779880
(1,    10001):      0.071979818344100
(1,   100001):      1.785329985620000
(1,  1000001):     54.977741956700000
(1, 10000001):   1700.231099130000000

It can go over a range of a million numbers in under a minute.

Usage

$ python test.py
21 41
3
$ python test.py
51 89
7
$ python test.py
201 501
3

Ungolfed

import collections

def is_it_prime(n):
    if n > 1:
        if all(n % i for i in range(int(n ** 0.5), 1, -1)):
            return True
    return False

start, stop = raw_input.split(' ', 1)

primes = ""
for p in range(int(start), int(stop)):
    if is_it_prime(p):
        primes += str(p)

counter = collections.Counter(primes).most_common(1)
print(counter[0][0])