After reading the papers by Rufus Isaacs [1] and George Spencer-Brown [2], I have reached to the conclusion that spiral chain edge coloring algorithm [3] gives answer to the question in affirmative.
[1] Rufus Isaacs, "Infinite families of nontrivial trivalent graphs which are not tait colorable", American Math Monthly 73 (1975) 221-239.
[2] George Spencer-Brown, "Uncolorable trivalent graphs", Thirteenth European Meeting on Cybernetics and Systems Research, University of Vienna, April 10, 1996.
[3] I. Cahit, Spiral Chains: The Proofs of Tait's and Tutte's Three-Edge-Coloring Conjectures. arXiv preprint, math CO/0507127 v1, July 6, 2005.
Let $e=uv$ be an edge of $G$.
Let $f$ and $g$ be the other two edges incident with $v$.
Let $G'=G-f-g$.
Claim 1: $G'-v$ is 2-connected.
$G'-v$ is the same as $G-v$.
Claim 2: $G'$ has a perfect matching.
You can prove this by showing directly that the Tutte condition holds.
I assume that you are familiar with the standard proof of Petersen's theorem,
so I will not elaborate on every detail.
Note that $G'$ has only 2 vertices of even degree.
Let $S$ be a vertex subset of $G$, $o(G'-S)$ the number of odd components of $G'-S$.
At most one odd component can contain $v$; that component can have as few as 1 edge to $S$ (since $G'$ is still connected).
All other odd components contain a vertex different from $v$, so because $G'-v$ is 2-connected all other odd components have at least 2 edges to $S$.
At most two odd components can contain a vertex of even degree (even in $G'$).
All other odd components must have at least three edges to $S$ (as in the
standard proof of the Petersen theorem).
Let $x$ be the number of edges between $S$ and the odd components of $G'-S$.
We have just shown that $x\geq 3o(G'-S)-4$.
On the other hand $S$ can accomodate at most $3|S|$ edges, so $x\leq3|S|$.
This means that $o(G'-S)\leq|S|+\frac{4}{3}$.
Now finally use the fact that $G$ is even, so $o(G'-S)$ and $|S|$ have the same parity.
This shows that $o(G'-S)\leq|S|$ and the Tutte condition holds.
Final argument: the perfect matching of $G'$ is a perfect matching of $G$ and it contains $e$,
since the degree of $v$ in $G'$ is 1.
Best Answer
Let $\mathrm{odd}(H)$ denote the number of odd components of the graph $H$.
Let $G$ be a connected cubic graph with no more than two cut-edges. Assume by way of contradiction that $G$ has no perfect matching.
Since the number of vertices of $G$ is even, if $G$ has no perfect matching then a maximum size matching in $G$ must miss at least two vertices and so, by theorem Tutte-Berge theorem, there is a subset $S$ of $V(G)$ such that $|S|\le\mathrm{odd}(G\setminus S)-2$. Each odd component of $G\setminus S$ has an odd number of vertices of odd degree and therefore each odd component of $G\setminus S$ is joined to $S$ by an odd number of edges. If this odd number is one, the corresponding edge must be a cut-edge. By our hypothesis it follows all but at most two components of $G\setminus S$ are joined to $S$ by at least three edges, hence the odd components are joined to $S$ by at least $$ 3(\mathrm{odd}(G\setminus S)-2)+2\ge3|S|+2 $$ edges. But the number of edges in $G$ ending on vertices in $S$ is at most $3|S|$, and we have arrived at a contradiction.
Remark: The Tutte-Berge theorem asserts that the number of vertices not covered by a matching of maximum size is $\max_{S\subseteq V(G)}\{\mathrm{odd}(G\setminus S)-|S|\}$