I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number $B_k$: it is precisely, the product of primes p for which $p-1\mid k$ (when $p-1\nmid k$, a result of Kummer says that $B_k/k$ is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of p-adic L-functions, specifically, for k a positive integer $$\zeta_p(1-k)=(1-p^{k-1})(-B_k/k),$$ where $\zeta_p$ is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example). For a totally real field F, the Deligne-Ribet p-adic L-function provides a generalization of the p-adic Riemann zeta function exists, namely the p-adic Dedekind zeta function $\zeta_{F,p}$ (as proved independently by Deligne–Ribet (Inv Math 59), Cassou-Noguès (Inv Math 51), and Barsky (1978)). One link between these and the Leopoldt conjecture is through the p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s = 1 des fonctions zêta p-adiques" (Inv Math 91): $$\lim_{s\rightarrow1}(s-1)\zeta_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R_phE_p}{w\sqrt{D}}$$ where h is the class number, $$E_p=\prod_{\mathfrak{p}\mid p}\left(1-\mathcal{N}(\mathfrak{p})^{-1}\right)$$ is a product of Euler-like factors, w = 2 is the number of roots of unity, D is the discriminant and $R_p$ is the interesting part here: the p-adic regulator (as Colmez notes, $\sqrt{D}$ and $R_p$ both depend on a choice of sign, but their ratio does not).
Theorem: The Leopoldt conjecture is equivalent to the non-vanishing of the p-adic regulator.
(For this, see, for example, chapter X of Neukirch-Schmidt-Wingberg's "Cohomology of number fields").
A clear consequence of this is that if $\zeta_{F,p}$ does not have a pole at s = 1, then the Leopoldt conjecture is false for (F, p). Perhaps an understanding of the denominators of values of $\zeta_{F,p}$ could lead to an understanding of the pole at s = 1 of $\zeta_{F,p}$.
Added (2010/04/09): So here's how you can use von Staudt–Clausen to see that the $p$-adic zeta function (of Q) has a pole at s = 1. It is clear from your statment of vS–C that it is saying that for $k\equiv0\text{ (mod }p-1)$, $B_k\equiv -1/p\text{ (mod }\mathbf{Z}_p)$ (i.e. it is not $p$-integral). Let $k_i=(p-1)p^i$, the $k_i$ is $p$-adically converging to 0, so $\zeta_p(1-k_i)$ is approaching $\zeta_p(1)$ (since $\zeta_p(s)$ is $p$-adically continuous, at least for $s\neq1$). By the aforementioned interpolation property of $\zeta_p(1-k)$, we have $$v_p(\zeta_p(1-k_i))=v_p(B_{k_i}/k_i)=-1-i\rightarrow -\infty$$ hence $1/\zeta_p(1-k_i)$ is approaching 0.
