Is there a map of spectra implementing the Thom isomorphism?

There is a construction for both Thom isomorphisms, homological and cohomological, via classical stable homotopy theory. You find the details in Rudyaks book "On Thom spectra, orientability, and cobordism", chapter V, §1. The Thom class is a map $X^{\mu} \to\Sigma^{n} H \mathbb{Z}$. Moreover, there is a map of spectra $X^{\mu} \to X_+ \wedge X^{\mu}$ which is induced from the map of vector bundles $\mu \to \mathbb{R}^0 \times \mu$ over the diagonal map $X \to X \times X$. Here is the definition of the homological Thom isomorphism; the cohomological one is in the same spirit. Consider the composition

$X^{\mu} \wedge H \mathbb{Z} \to X_+ \wedge X^{\mu} \wedge H\mathbb{Z} \to X_+ \wedge \Sigma^n H \mathbb{Z} \wedge H \mathbb{Z} \to X_+ \wedge \Sigma^n H \mathbb{Z} $. On homotopy groups, it induces a map lowering the degree by $n$ (there is a sign mistake in your question that confused me for some minutes).

It is clear that this works for orientations with respect to other ring spectra as well.

Johannes points out that it works quite generally, and perhaps it is useful to see that generality. Pardon me for going back to basics here. There are three ingredients:

1. Given an n-plane bundle $\xi \to B$ we have the Thom diagonal $T(\xi) \to T(\xi) \wedge B_+ $ induced by the diagonal map $B_+ \to B_+ \wedge B_+$, expressing $T(\xi)$ as a comodule over $B_+$ and the natural map $B_+ \to T(\xi)$ as a comodule map.

2. A ring spectrum $E$.

3. An orientation $T(\xi) \to \Sigma^n E$. This is a cohomology class in $E^n T(\xi)$ which restricts to a generator of $\pi_*E$ when composed with the natural map $S^n \to T(\xi)$ induced by the inclusion of a point into $B$. (See pp. 253-255 (following Prop. 10.5) of Adams' "Stable Homotopy and Generalised Homology" for a nice discussion of why this is the right way to define an orientation.)

The `geometric' Thom isomorphism is then the homotopy equivalence obtained by composing these:

$E \wedge T(\xi) \to E \wedge T(\xi) \wedge B_+ \to E \wedge \Sigma^n E \wedge B_+ \to E \wedge \Sigma^n B_+$

Applying $\pi_* $ we obtain the Thom isomorphism $E_* T(\xi) \cong E_* \Sigma^n B_+$.

Since it is an $E$-module map, we can apply $F_E(-,E)$ and the equivalence $F_E(E\wedge X,E) \simeq F(X,E)$ to get an equivalence $F(\Sigma^n B_+,E) \to F(T(\xi),E)$ inducing the Thom isomorphism in $E$-cohomology, $E^*\Sigma^n B_+ \cong E^* T(\xi) $.

I just noticed this question, so my apologies for a very belated answer. Proposition 20.5.5 of http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf states that a $k$-orientation of a spherical fibration $X$ over a base space $B$ specifies a $k$-trivialization of $X$. Here $k$ is any (homotopy) commutative ring spectrum, and the notion of a $k$-orientation is the standard one. However, a $k$-trivialization is an equivalence $k\wedge X\simeq k\wedge S^n_B$ of parametrized spectra over $B$. Intuitively, after smashing with $k$ over $B$, $X$ thinks that it is the trivial fibration $B \times S^n \to B$.The Thom isomorphism follows directly.