No, not at all.
Consider the following example which is superficially closely related to Simon's algorithm and Shor's algorithm. Suppose you have two functions f and g, from ${\bf Z}_{2^N}$ to $\{0,1\}^m$ with $m\geqslant N$. Suppose they are random oracles with the property that $f(x+s \text{ mod } 2^N) = g(x)$ for some random s picked uniformly from ${\bf Z}_{2^N}$, and that if $x\neq y$, $f(x)\neq f(y)$. Otherwise, the functions are picked uniformly at random.
Can a quantum computer figure out s in polynomial time? Based on the superficial similarity with Simon's and Shor's algorithms, you might think so. Have three registers: the first a qubit, the second N qubits specifying an element of ${\bf Z}_{2^N}$, and the third giving the function value. It's easy to prepare the state $$\frac{1}{2^{(N+1)/2}} \sum_{x\in {\bf Z}_{2^N}} \left[ |0\rangle |x\rangle|f(x)\rangle +|1\rangle |x\rangle |g(x)\rangle \right].$$
Then, take the quantum Fourier transform of the second register, and the Hadamard transform of the first register. Then, measure the values of the first qubit and the second register in the computational basis. Repeat this $cN$ times where c is around 4 or so.
Each measurement, the measured value for the second register, k, is uniformly distributed over ${\bf Z}_{2^N}$. With probability $p_0(k,s)\equiv \cos^2(\pi ks/2^N)$, the first register will measure 0, and with probability $p_1(k,s)\equiv \sin^2(\pi ks/2^N)$, it will measure 1.
So, we have the pairs $(r_i,k_i)$. With probability nearly one, this sequence of pairs already contain enough information to figure out the value for $s$. Just evaluate the function $h(x)\equiv \prod_i p_{r_i}(k,x)$. $h(s)$ will have the largest value by far, typically exceeding a certain bound with nearly probability 1, while the other values will typically be less than another bound with a probability of nearly 1.
If $s\neq 0$, this can be found out the moment the r value of 1 shows up. Similarly, if $s$ isn't an integer multiple of $2^{N-a}$, this can be found out after an order of $2^a$ measurements or so. However, this can also be verified by $2^a$ direct evaluations.
The only problem is there is no known polynomial time quantum algorithm which can extract the value of s given these pairs. So, this algorithm can output a sequence of pairs which can be used to test if a number x is s, or not, but it can't find s by itself.
If there is a quantum algorithm which could, then the closely related problem of finding the seed of the pseudorandom generator can also likely be solved. p is a prime number N bits long. The seed is $y_0\in Z_p$. $y_{i+1}=ay_i +b \text{ mod p}$. $x_i$ is the last bit of $y_i$. Given the sequence $x_i$ for i up to $cN$, figure out the seed.
$y_i=[y_0+b/(a-1)]a^i-b/(a-1) \text { mod p}$