Common practise is to compare p-value with three levels - 0.05, 0.01 and 0.001. Since your p-value is less than each of them, you have to choose the smallest one, so you should conclude that differences are significant and p<0.001. Roughly speaking: The smaller the p-value, the more significant differences are.
Since we do not know distribution of your data, we do not also know which test should you use. But you have quite large sample, so there is high chance that parametric test can be appropriate (t-test for paired data).
This largely follows on from Antoni's answer.
Here's what I was getting at with a permutation test
First let's exclude the tie as Antoni did:
d s r
10 1 6
-9 -1 5
8 1 4
5 1 2
-6 -1 3
0 NA NA
2 1 1
Our test statistic is then 1+2+4+6 = 13
So we could calculate the permutation distribution from scratch as follows:
pm <- c(-1,1)
sign <- expand.grid(s1=pm,s2=pm,s3=pm,s4=pm,s5=pm,s6=pm)
table(rowSums((sign>0)%*%cbind(1:6)))/64
which gives the same values as dsignrank(0:21,6) (yay!). So clearly when there are no ties (which is the case, since we excluded them) this is just the signed rank test, as it should be
Let's consider whether we can do anything with the tie. How we really deal with it depends on how it came about.
If, for example we envision an underlying continuous scale that has become discretized into categories, then our tie is simply due to two different values that our subsequently imposed categories are not fine enough to distinguish.
Arguably then, the two values are "really" different but we have no basis to conclude which sign we should have and so allowing for both possibilities, half the time giving it a +1 and half the time giving it a -1 and then averaging the two (or perhaps to be conservative, some would argue we should take the higher of the two possible counts ... but I'll continue to look at the average of the two possible cases here). [By contrast if the values were inherently discrete with no underlying-but-unobserved continuous scale, then we would say that the 0 was inherent and deal with it in a different way -- I think exclusion probably makes the most sense in that case]
Note tie will always have the smallest rank.
So we have a table like the one above but the second last row is either:
d s r
: : :
0 -1 1
or
d s r
: : :
0 1 1
and all the other rows would have r one higher than before.
d s r
10 1 7
-9 -1 6
8 1 5
5 1 3
-6 -1 4
0 ±1 1
2 1 2
So the possible test statistics are 2+3+5+7 and 1+2+3+5+7 (17 and 18)
[One alternative approach, if we're being conservative would be to treat it as a full n=7 and take the larger of the two possible p-values which gives 0.6875]
If we do all the permutations but always average the two signs for the smallest rank, they'll contribute 1/2, and each of the other positive ranks will be 1 higher.
In effect we'd be doing this:
table(0.5+rowSums((sign>0)%*%cbind(2:7)))/64
And comparing a test statistic of 17.5 with that. This gives p = 0.625.
So anyway, all the approaches are a bit different but they pretty much tell the same general story here.
Best Answer
You just described the difference. No one can know in advance outcome differences because it greatly depends on the nature of your data.
Do you know the non-normal distribution you're working with? If so, you could simulate some results and see what the typical error rates for the different tests were and how they differed.