Java 10, 307 298 288 280 273 271 bytes

s->{String[]a=s.split("\\d"),o=a[1].split("\n"),f=a[2].split("\n");int r[]=new int[f.length-1],q=-2,d,i,j,z;for(var p:f)if(++q>=0)for(z=d=o.length,i=0;++i<d;)for(j=0;o[i].contains((f=p.split(" "))[0])&++j<z;)r[q]=d=o[j].contains(f[1])?Math.min(i>j?i-j:j-i,d):d;return r;}

-19 bytes thanks to @ceilingcat.

Explanation:

Try it here.

s->{                           // Method with String parameter & integer-array return
  String[]a=s.split("\\d"),    //  Split the input on numbers
          o=a[1].split("\n"),  //  List of orchards
          f=a[2].split("\n");  //  List of pairs (later on reused for individual pair of fruits)
  int r[]=new int[f.length-1], //  Result integer-array
      q=-2,d,i,j,z;            //  Some temp integers
  for(var p:f)                 //  Loop `p` over the pairs:
    if(++q>=0){                //   If this isn't the first iteration of the pair-loop
                               //   (the first item is always empty after `.split("\\d")`)
      for(z=d=o.length,i=0;++i<d;)
                               //    Inner loop `i` over the orchards:
        for(j=0;o[i].contains((f=p.split(" "))[0])
                               //     If the current orchard contains the first fruit of the pair:
                &++j<z;)       //      Inner loop `j` over the orchards again:
          r[q]=                //       Add the following to the resulting array:
               d=o[j].contains(f[1])?
                               //       If the current orchard contains the second fruit of the pair:
                 Math.min(     //        Take the minimum of:
                   i>j?i-j:j-i,//         The absolute difference between `i` and `j`
                   d)          //         And the current integer of the array
               :               //       Else:
                d;             //        Leave the current integer unchanged
  return r;}                   //  Return the result array

Röda, 118 bytes

{A=[head(parseInteger(pull()))|splitMany|enum]pull n;split|{f={|x|A|[_2]if[x in _1]}f a|{|i|f b|abs i-_}_|min}for a,b}

Explanation:

{
    A=[
        /* Pull line, convert to integer, and read that many lines */
        head(parseInteger(pull()))|
        /* Split all lines to arrays */
        splitMany|
        /* Enumerate, ie. give each line a number */
        enum
    ]
    /* Discard one line */
    pull n;
    /* Split each following line and push words to the stream */
    split|
    /* For each word a, b in the stream: */
    {
        /* Helper function to find lines that contain x */
        f={|x|A|[_2]if[x in _1]}
        /* Push numbers of lines that contain the first word to the stream */
        f a|
        /* For each line number i */
        {|i|
            /* Push numbers of lines that contain the second word to the stream */
            f b|
            /* Push the difference of two line numbers to the stream */
            abs i-_
        }_|
        /* Push the smallest difference to the stream */
        min
    }for a,b
}

No TIO link yet, as the TIO version of Röda is outdated.

JavaScript (ES6), 224 bytes

a=>(d=a.split('\n')).slice(e=+d[0]+2).map(f=>{p=e;for(g=(q=(t,s=0)=>d.slice(1,e-1).map(b=c=>c.split(' ')).findIndex((x,y)=>y>=t&x.includes(b(f)[s])))(0);++g;g=q(g))for(l=q(0,1);++l;l=q(l,1))p=p<(u=g>l?g-l:l-g)?p:u;return p})

Call with "6\nnone\napple\norange pear pear\nnone\norange lemon pumpkin\npumpkin lettuce flowers peas\n4\npeas lettuce\napple orange\napple pumpkin\nflowers orange"

176 bytes, Python 3

def p(a):l=[b.split()for b in a.split('\n')];X=int(l[0][0]);return[min(y-x for y in range(X)for x in range(y+1)if z[0]in l[x+1]+l[y+1]and z[1]in l[x+1]+l[y+1])for z in l[X+2:]]

This can be called by passing in the entire content as a string

p("""6
none
apple
orange pear pear
none
orange lemon pumpkin
pumpkin lettuce flowers peas
4
peas lettuce
apple orange
apple pumpkin
flowers orange""")

Ungolfed solution

def problem(a):
    # this will split on new line, and split each individual line into it's own list
    lines=[b.split()for b in a.split('\n')]
    # get the number of orchards
    X=int(lines[0][0])
    return[
        # get the minimum of all possible differences
        min(y-x # subtract y and x to get the difference
            for y in range(X) # for each orchard, y
                for x in range(y+1) # and each orchard up to y inclusive, x
                    if z[0] in lines[x+1]+lines[y+1] and z[1] in lines[x+1]+lines[y+1]) # if both fruits exist the x and y orchards
        # for every day (z is a list containing 2 fruits as string)
        for z in lines[X+2:]
    ]