Pearson's r and Spearman's rho are both already effect size measures. Spearman's rho, for example, represents the degree of correlation of the data after data has been converted to ranks. Thus, it already captures the strength of relationship.
People often square a correlation coefficient because it has a nice verbal interpretation as the proportion of shared variance. That said, there's nothing stopping you from interpreting the size of relationship in the metric of a straight correlation.
It does not seem to be customary to square Spearman's rho. That said, you could square it if you wanted to. It would then represent the proportion of shared variance in the two ranked variables.
I wouldn't worry so much about normality and absolute precision on p-values. Think about whether Pearson or Spearman better captures the association of interest. As you already mentioned, see the discussion here on the implication of non-normality for the choice between Pearson's r and Spearman's rho.
The Spearman's rank c. c. is the Pearson' c.c. of the ranked variables; in its turn the Pearson's c.c. is defined as the mean of the product of the paired standardized scores $z(X_i)$, $z(Y_i)$.
\begin{equation}
r(X,Y) = \Sigma_i[z(X_i) z(Y_i)]/(n-1)
\end{equation}
in which $n$ is the sample size and the standard scores
\begin{equation}
z(X_i) = [X_i - \bar{X}]/std(X)
\end{equation}
\begin{equation}
z(Y_i) = [Y_i - \bar{Y}]/std(Y)
\end{equation}
are relative to the ranked variables ($X_i$, $Y_i$). Squaring $r(X_i, Y_i)$ we obtain the coefficient of determination $r²$, which we can equate to the fraction of explained variance. So if my Spearman's rank c.c. is of 0.6, I can deduce that the variance of the ranked variables is shared at 36%.
From the first equation and attempting at a simpler way of explaining $r(X,Y)$, I would say is the average value of concordance of z-score variations. For instance, let us say I repeat an experiment by increasing the sample size $n$ and calculate $r(X,Y)$ for both the small sample and the larger one. Let us say that associated to an increase in n of $~3$ I get a decrease in $r(X,Y)$ of roughly 50%; this corresponds to a decrease in standard scores concordance of 50%. My interpretation should be then that the latest dataset provides weaker evidence for the presence of a correlation in the data.
Best Answer
While ranking the data for use in Spearman correlation is possible with Excel formulas (like almost everything), it is not that easy.
I would suggest a little easier solution, that at the moment will work only in 32-bit Excel: use RExcel:
If you want to have the Spearman correlation coefficient $\rho$, type in this formula:
=REval("cor.test(var1,var2,method='spearman')$estimate")